AP Inter 2nd Year Maths Exercise 7e Solutions

Referring to the AP Inter 2nd Year Maths Study Material Chapter 7 Integrals Exercise 7e Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Integrals Solutions Exercise 7e

I.

Question 1.
Find the integral of \(\frac{x}{(x+1)(x+2)}\)
Solution:
Let \(\frac{\mathrm{x}}{(\mathrm{x}+1)(\mathrm{x}+2)}=\frac{\mathrm{A}}{(\mathrm{x}+1)}+\frac{\mathrm{B}}{(\mathrm{x}+2)}=\frac{\mathrm{A}(\mathrm{x}+2)+\mathrm{B}(\mathrm{x}+1)}{(\mathrm{x}+1)(\mathrm{x}+2)}\)
⇒ A(x + 2) + B(x + 1) = x …………(1)
Put x = -1 in (1) ⇒ A(1) + 0 = -1 ⇒ A = -1
Put x = -2 in (1) ⇒ 0 + B(-1) = -2 ⇒ B = 2
\(\frac{x}{(x+1)(x+2)}=\frac{-1}{(x+1)}+\frac{2}{(x+2)}\)
⇒ \(\int \frac{x}{(x+1)(x+2)} d x=\int \frac{-1}{(x+1)}+\frac{2}{(x+2)} d x\) = -log |x + 1| + 2 log|x + 2| + C
= log(x + 2)² – log(x + 1) + C = \(\log \frac{(x+2)^2}{|x+1|}+C\) [∵ loga – logb = \({log}\left(\frac{\mathrm{a}}{\mathrm{~b}}\right)\)]

Question 2.
Find the integral of \(\frac{1}{x^2-9}\)
Solution:
Let \(\frac{1}{x^2-9}=\frac{1}{(x+3)(x-3)}=\frac{A}{(x+3)}+\frac{B}{(x-3)}=\frac{A(x-3)+B(x+3)}{(x+3)(x-3)}\)
⇒ A(x – 3) + B(x + 3) = 1 …………..(1)
Put x = 3 in (1) ⇒ 1 = A(3 – 3) + B(3 + 3) ⇒ 6B = 1 ⇒ B = 1/6
Put x = -3 in (1) ⇒ 1 = A(-3 – 3) + B(-3 + 3) ⇒ -6A = 1 ⇒ A = -1/6
∴ \(\frac{1}{(x+3)(x-3)}=\frac{-1}{6(x+3)}+\frac{1}{6(x-3)}\)
⇒ \(\int \frac{1}{\left(x^2-9\right)} d x=\int\left(\frac{-1}{6(x+3)}+\frac{1}{6(x-3)}\right) d x\)
= \(-\frac{1}{6} \log |x+3|+\frac{1}{6} \log |x-3|+C=\frac{1}{6} \log \left|\frac{x-3}{x+3}\right|+C\)

Question 3.
Find the integral of \(\frac{1-x^2}{x(1-2 x)}\)
Solution:
The given integral is an improper rational function.
So by dividing the numerator 1 – x² by the denominator x(1 – 2x) = x – 2x² we get
\(\frac{1-x^2}{x(1-2 x)}=\frac{1}{2}+\frac{1}{2}\left(\frac{2-x}{x(1-2 x)}\right)\)
Now let \(\frac{2-x}{x(1-2 x)}=\frac{A}{x}+\frac{B}{1-2 x}=\frac{A(1-2 x)+B x}{(x)(1-2 x)}\)
A(1 – 2x) + Bx = (2 – x) ……..(1)
Pit x = 0 in (1) ⇒ A(1 – 0) + B(0) = 2 – 0 ⇒ A = 2
Put x = 1/2 in (1) ⇒ A(0) + B(1/2) = 1 – 0 ⇒ 2 – \(\frac{1}{2}=\frac{3}{2}\)
∴ \(\frac{2-x}{x(1-2 x)}=\frac{2}{x}+\frac{3}{1-2 x}\)
∴ \(\frac{1-x^2}{x(1-2 x)}=\frac{1}{2}+\frac{1}{2}\left[\frac{2}{x}+\frac{3}{1-2 x}\right] \Rightarrow \int \frac{1-x^2}{x(1-2 x)} d x=\int\left[\frac{1}{2}+\frac{1}{2}\left(\frac{2}{x}+\frac{3}{1-2 x}\right)\right] d x\)
= \(\frac{x}{2}+\log |x|+\frac{3}{2(-2)} \log |1-2 x|+C\) [∵ \(\int \frac{1}{a x+b} d x=\frac{1}{a} \log |a x+b|+C\)]
= \(\frac{x}{2}+\log |x|-\frac{3}{4} \log |1-2 x|+C\)

II.

Question 1.
Find the integral of \(\frac{3 x-1}{(x-1)(x-2)(x-3)}\)
Solution:
Let \(\frac{3 x-1}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}\)
= \(\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)}\)
⇒ A(x – 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x – 2) = 3x – 1 …..(1)
Put x = 1 in (1) ⇒ A(-1)(-3) + 0 + 0 = 3 – 1 ⇒ 2A = 2 ⇒ A = 1
Put x = 2 in (1) ⇒ 0 + B1(-1) + 0 = 6 – 1 ⇒ -B = 5 ⇒ B = -5
Put x = 3 in (1) ⇒ 0 + 0 + C(2)1 = 9 – 1 ⇒ 2C = 8 ⇒ C = 4
∴ \(\frac{3 x-1}{(x-1)(x-2)(x-3)}=\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}\)
⇒ \(\int \frac{3 x-1}{(x-1)(x-2)(x-3)} d x=\int\left[\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}\right] d x\)
= log |x – 1| – 5log |x – 2| + 4log |x – 3| + C

Question 2.
Find the integral of \(\frac{x}{(x-1)(x-2)(x-3)}\)
Solution:
Let \(\frac{\mathrm{x}}{(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)}=\frac{\mathrm{A}}{(\mathrm{x}-1)}+\frac{\mathrm{B}}{(\mathrm{x}-2)}+\frac{\mathrm{C}}{(\mathrm{x}-3)}\)
= \(\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)}\)
A(x – 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x -2) = x ……..(1)
Put x = 1 in (1) ⇒ A(-1)(-2) + 0 + 0 = 1 ⇒ 2A = 1 ⇒ A = 1/2
Put x = 2 in (1) ⇒ 0 + B1(-1) + 0 = 2 ⇒ -B = 2 ⇒ B = -2
Put x = 3 in (1) ⇒ 0 + 0 + C(2)1 = 3 ⇒ 2C = 3 ⇒ C = 3/2
∴ \(\frac{x}{(x-1)(x-2)(x-3)}=\frac{1}{2(x-1)}-\frac{2}{(x-2)}+\frac{3}{2(x-3)}\)
⇒ \(\begin{aligned}
\int \frac{x}{(x-1)(x-2)(x-3)} d x=\int & {\left[\frac{1}{2(x-1)}-\frac{2}{(x-2)}+\frac{3}{2(x-3)}\right] d x } \\
& =\frac{1}{2} \log |x-1|-2 \log |x-2|+\frac{3}{2} \log |x-3|+C
\end{aligned}\)

Question 3.
Find the integral of \(\frac{2 x}{x^2+3 x+2}\)
Solution:
Let \(\frac{2 x}{x^2+3 x+2}=\frac{2 x}{(x+1)(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}=\frac{A(x+2)+B(x+1)}{(x+1)(x+2)}\)
⇒ A(x + 2) + B(x + 1) = 2x …….(1)
Put x = -1 ⇒ A(-1 + 2) + B(0)2(-1) ⇒ A = -2
Put x = -2 ⇒ A(-2 + 2) + B(-2 + 1) = 2(-2) ⇒ B = -4
∴ \(\frac{2 x}{(x+1)(x+2)}=\frac{-2}{(x+1)}+\frac{4}{(x+2)} \Rightarrow \int \frac{2 x}{(x+1)(x+2)} d x=\int\left[\frac{4}{(x+2)}-\frac{2}{(x+1)}\right] d x\)
= 4log |x + 2| – 2log |x + 1| + C

Question 4.
Find the integral of \(\frac{x}{\left(x^2+1\right)(x-1)}\)
Solution:
Let \(\frac{x}{\left(x^2+1\right)(x-1)}=\frac{A x+B}{\left(x^2+1\right)}+\frac{C}{(x-1)}=\frac{(A x+B)(x-1)+C\left(x^2+1\right)}{\left(x^2+1\right)(x-1)}\)
⇒ (Ax + B)(x – 1) + C(x² + 1) = x ………….(1)
Put x = 1 in (1) ⇒ 0 +C(1 + 1) = 1 ⇒ 2C = 2 ⇒ C = 1/2
Put x = 0 in (1) ⇒ (0 + B)( 0 – 1) + C(0 + 1) = 0 ⇒ -B + C = 0 ⇒ B = C = 1/2
Equating the coefficients of x²
⇒ A + C = 0 ⇒ A = -C = -1/2 A = \(-\frac{1}{2}\), B = \(\frac{1}{2}\) and C = \(\frac{1}{2}\)

Question 5.
Find the integral of \(\frac{2}{(1-x)\left(1+x^2\right)}\)
Solution:
Let \(\frac{2}{(1-x)\left(1+x^2\right)}=\frac{A}{(1-x)}+\frac{B x+C}{\left(1+x^2\right)}=\frac{A\left(1+x^2\right)+(B x+C)(1-x)}{(1-x)\left(1+x^2\right)}\)
⇒ A(1 + x²) + (Bx + C)(1 – x) = 2 ……..(1)
Put x = 1 in (1) ⇒ A(1 + 1) + 0 = 2 ⇒ 2A = 2 ⇒ A = 1
Equating the coefficients of x² we get A – B = 0 ⇒ B = A = 1
Equating the coefficients of x we get B – C = 0 ⇒ C = A = 1
∴ \(\frac{2}{(1-x)\left(1+x^2\right)}=\frac{1}{1-x}+\frac{x+1}{1+x^2}\) [\(\int \frac{f^{\prime}(x)}{f(x)} d x\) = log |f(x)| + c]
⇒ \(\int \frac{2}{(1-x)\left(1+x^2\right)} d x=\int \frac{1}{1-x} d x+\int \frac{x}{1+x^2} d x+\int \frac{1}{1+x^2} d x\)
= \(-\int \frac{-1}{x-1} d x+\frac{1}{2} \int \frac{2 x}{1+x^2} d x+\int \frac{1}{1+x^2} d x=-\log |x-1|+\frac{1}{2} \log \left(1+x^2\right)+\tan ^{-1} x+C\)

Question 6.
Find the integral of \(\frac{3 x-1}{(x+2)^2}\)
Solution:
Let \(\frac{3 x-1}{(x+2)^2}\)
Consider \(\frac{3 x-1}{(x+2)^2}\)
Put x + 2 = y ⇒ x = y – 2
\(\begin{aligned}
\frac{3 x-1}{(x+2)^2} & =\frac{3(y-2)-1}{(y)^2} \\
& =\frac{3 y-6-1}{y^2}=\frac{3 y-7}{y^2} \\
& =\frac{3}{y}-\frac{7}{y^2}
\end{aligned}\)
⇒ A(x + 2) + B = 3x – 1 …………….(1)
Put x = -2 in (1) ⇒ A(0) + B = 3(-2) – 1 ⇒ B = 7
Comparing the coefficient of x we get A = 3

Question 7.
Find the integral of \(\frac{1}{x\left(x^n+1\right)}\)
[Hint: Multiply numerator and denominator by x^n-1 and put xⁿ = t]
Solution:
Multiplying numerator and denominator of the integral by x^n-1, we get
\(\frac{1}{x\left(x^n+1\right)}=\frac{x^{n-1}}{x^{n-1} x\left(x^n+1\right)}=\frac{x^{n-1}}{x^n\left(x^n+1\right)}\)
Put xⁿ = t ⇒ nx^n-1dx = dt
∴ \(\int \frac{1}{x\left(x^n+1\right)} d x=\int \frac{x^{n-1}}{x^n\left(x^n+1\right)} d x=\frac{1}{n} \int \frac{1}{t(t+1)} d t\)
Let \(\frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{(t+1)}=\frac{A(1+t)+B t}{t(t+1)}\)
A(1 + t) + Bt = 1 …………(1)
Put t = 0 in (1) ⇒ A(1 + 0) + 0 = 1 ⇒ A = 1
Put t = -1 in (1) ⇒ 0 – B = 1 ⇒ B = -1

Question 8.
Find the integral of \(\frac{1}{\left(e^x-1\right)}\) [Hint: Put e^x = t]
Solution:
Put e^x = t ⇒ e^x dx = dt
∴ \(\int \frac{1}{\left(e^x-1\right)} d x=\int\left(\frac{1}{t-1}\right) \frac{d t}{t}=\int \frac{1}{t(t-1)} d t\)
Let \(\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}=\frac{A(t-1)+B t}{t(t-1)}\) ⇒ A(t – 1) + Bt = 1 ………..(1)
Put t = 0 in (1) ⇒ A(0 – 1) + 0 = 1 ⇒ A = -1
Put t = 1 in (1) ⇒ 0 – B = 1 ⇒ B = 1
∴ \(\frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1}\) [∵ log a – log b = \(\log \left(\frac{a}{b}\right)\)]
⇒ \(\int \frac{1}{t(t-1)} d t=-\log t+\log |t-1|=\log \left|\frac{t-1}{t}\right|+C=\log \left(\frac{e^x-1}{e^x}\right)+C\)

Question 9.
Find the integral of \(\frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)}\)
Solution:
Put e^x = t ⇒ e^x dx = dt
∴ \(\int \frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)} d x=\int \frac{d t}{(t+1)(t+2)}=\int \frac{(t+2)-(t+1)}{(t+1)(t+2)} d t=\int\left[\frac{1}{(t+1)}-\frac{1}{(t+2)}\right] d t\)
= log |t + 1| – log |t + 2| + C = \(\log \left|\frac{t+1}{t+2}\right|+C=\log \left|\frac{1+e^x}{2+e^x}\right|+C\)

Question 10.
Find the integral of \(\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}\)
Solution:
Put x² = t.
Then \(\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}=\frac{1}{(t+1)(t+4)}=\frac{A}{(t+1)}+\frac{B}{(t+4)}=\frac{A(t+4)+B(t+1)}{(t+1)(t+4)}\)
⇒ A(t + 4) + B(t + 1) = 1 ………(1)
Put t = -1 in (1) ⇒ A(-1 + 4) + B(0) = 1 ⇒ 3A = 1 ⇒ 3A = 1/3
Put t = -4 in (1) ⇒ A(0) + B(-4 + 1) = 1 ⇒ -3B = 1 ⇒ B = -1/3

III.

Question 1.
Find the integral of \(\frac{x}{(x-1)^2(x+2)}\)
Solution:
Let \(\frac{x}{(x-1)^2(x+2)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+2)}=\frac{A(x-1)(x+2)+B(x+2)+C(x-1)^2}{(x-1)^2(x+2)}\)
A(x – 1)(x + 2) + B(x + 2) + C(x – 1)² = x …….(1)
Put x = 1 in (1) ⇒ 0 + 3B + 0 = 1 ⇒ 3B = 1 ⇒ B = 1/3
Put x = -2 in (1) ⇒ 0 + 0 + C(-2 – 1)² = -2 ⇒ 9C = -2 ⇒ C = -2/9
Equating the coefficient of x² we get
A + C = 0 ⇒ A = -C = 2/9 A = \(\frac{2}{9}\), B = \(\frac{1}{3}\) and C = \(-\frac{2}{9}\)

Question 2.
Find the integral of \(\frac{3 x+5}{x^3-x^2-x+1}\)
Solution:
We have \(\frac{3 x+5}{x^3-x^2-x+1}=\frac{3 x+5}{(x-1)^2(x+1)}\)
Let \(\frac{3 x+5}{(x-1)^2(x+1)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+1)}=\frac{A(x-1)(x+1)+B(x+1)+C(x-1)^2}{(x-1)^2(x+2)}\)
⇒ A(x – 1)(x + 1) + B(x + 1) + C(x – 1)² = 3x + 5 ….(1)
Put x = 1 in (1) ⇒ 0 + 2B + 0 = 3 + 5 ⇒ B = 4
Put x = -1 in (1) ⇒ 0 + 0 + 4C = -3 + 5 ⇒ 4C = 2 ⇒ C = 1/2
Equating the coefficient of x² we get
A + C = 0 ⇒ A = -C = -1/2 ∴ A = \(-\frac{1}{2}\), B = 4 and C = \(\frac{1}{9}\)

Question 3.
Find the integral of \(\frac{2 x-3}{\left(x^2-1\right)(2 x+3)}\)
Solution:
We have \(\frac{2 x-3}{\left(x^2-1\right)(2 x+3)}=\frac{2 x-3}{(x-1)(x+1)(2 x+3)}\)
Let \(\frac{2 x-3}{(x-1)(x+1)(2 x+3)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{C}{(2 x+3)}\)
= \(\frac{A(x-1)(2 x-3)+B(x+1)(2 x+3)+C(x+1)(x-1)}{(x+1)(x-1)(2 x-3)}\)
⇒ A(x – 1)(2x – 3) + B(x + 1)(2x + 3) + C(x + 1)(x – 1) = 2x – 3 …..(1)
Put x = -1 in (1) ⇒ A(-1 – 1)(-2 + 3) + 0 + 0 = -2 – 3 ⇒ -2A = -5 ⇒ A = 5/2
Put x = 1 in (1) ⇒ 0 + B(2)(5) + 0 = 2 – 3 ⇒ 10B = -1 ⇒ B = -1/10
Put x = -3/2 in (1) ⇒ 0 + 0 + C\(\left(\frac{-3}{2}+1\right)\left(\frac{-3}{2}-1\right)=2\left(-\frac{3}{2}\right)-3 \Rightarrow-6=C\left(\frac{-1}{2}\right)\left(\frac{-5}{2}\right)\)
⇒ 5C = -24 ⇒ C = \(-\frac{24}{5}\)
∴ A = \(\frac{5}{2}\), B = \(-\frac{1}{10}\) and C = \(-\frac{24}{5}\)

Question 4.
Find the integral of \(\frac{5 x}{(x+1)\left(x^2-4\right)}\)
Solution:
We have \(\frac{5 x}{(x+1)\left(x^2-4\right)}=\frac{5 x}{(x+1)(x+2)(x-2)}\)
Let \(\frac{5 x}{(x+1)\left(x^2-4\right)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}+\frac{C}{(x-2)}\)
= \(\frac{A(x+2)(x-2)+B(x+1)(x-2)+C(x+1)(x+2)}{(x+1)(x+2)(x-2)}\)
A(x + 2)(x – 2) + B(x + 1)(x – 2) + C(x+ 1)(x + 2) = 5x ……..(1)
Put x = -1 in (1) ⇒ A(-1 + 2)(-1 – 2) + 0 + 0 = -5 ⇒ -3A = -5 ⇒ A = 5/3
Put x = -2 in (1) ⇒ 0 + B(-2 + 1)(-2 – 2)(5) + 0 = -10 ⇒ 4B = -10 ⇒ B = -5/2
Put x = 2 in (1) ⇒ 0 + 0 + C(2 + 1)(2 + 2) = 10 ⇒ 12C = 10 ⇒ C = 5/6
A = \(\frac{5}{3}\), B = \(-\frac{5}{2}\) and C = \(\frac{5}{6}\)

Question 5.
Find the integral of \(\frac{x^3+x+1}{x^2-1}\)
Solution:
The given integral is an improper rational function
On dividing (x³ + x + 1) by x² – 1, we get \(\frac{x^3+x+1}{x^2-1}=x+\frac{2 x+1}{x^2-1}\)
Let \(\frac{2 x+1}{x^2-1}=\frac{A}{(x+1)}+\frac{B}{(x-1)}=\frac{A(x-1)+B(x+1)}{x^2-1}\)
⇒ A(x – 1) + B(x + 1) = 2x + 1 ………..(1)
Put x = -1 in (1) ⇒ A(-1 – 1) + 0 = -2 + 1 ⇒ -2A = -1 ⇒ A = 1/2
Put x = 1 in (1) ⇒ 0 + B(1 + 1) = 2 + 1 ⇒ 2B = 3 ⇒ B = 3/2

Question 6.
Find the integral of \(\frac{1}{x^4-1}\)
Solution:
GE = \(\frac{1}{\left(x^4-1\right)}=\frac{1}{\left(x^2-1\right)\left(x^2+1\right)}=\frac{1}{(x+1)(x-1)\left(x^2+1\right)}\)
Let \(\frac{1}{(x+1)(x-1)\left(x^2+1\right)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{C x+D}{\left(x^2+1\right)}\)
= \(\frac{A(x-1)\left(1+x^2\right)+B(x+1)\left(1+x^2\right)+(C x+D)\left(x^2-1\right)}{(x+1)(x-1)\left(x^2+1\right)}\)
A(x – 1)(1 + x²)
⇒ A(x³ + x – x² – 1) + B(x³ + x + x² + 1) + Cx³ + Dx² – Cx – D = 1
⇒ (A + B + C)x³ + (-A + B + D)x² + (A + B – C)x + (-A + B – D) = 1
Put x = -1 in(1) ⇒ A(-1 – 1)(1 + 1) + 0 + 0 = 1 ⇒ -4A = 1 ⇒ A = -1/4
Put x = 1 in (1) ⇒ 0 + B(1 + 1)(1 + 1) + 0 = 1 ⇒ 4B = 1 ⇒ B = 1/4
Equating the coefficient of x³ we get
A + B + C = 0 ⇒ C = -A – B = \(\frac{1}{4}-\frac{1}{4}\) = 0
Equating the coefficient of x² we get
-A + B – D = 1 ⇒ D = A – B = \(-\frac{1}{4}-\frac{1}{4}=\frac{-2}{4}=-\frac{1}{2}\)
∴ A = \(-\frac{1}{4}\), B = \(\frac{1}{4}\), C = 0, D = \(-\frac{1}{2}\)

Question 7.
Find the integral of \(\frac{\cos x}{(1-\sin x)(2-\sin x)}\) [Hint: Put sin x = t]
Solution:
Put sin x = t ⇒ cos xdx = dt
∴ I = \(\int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x=\int \frac{d t}{(1-t)(2-t)}\)
Let \(\frac{1}{(1-t)(2-t)}=\frac{A}{(1-t)}+\frac{B}{(2-t)}=\frac{A(2-t)+B(1-t)}{(2-t)(1-t)}\)
A(2 – t) + B(1 – t) = 1 …….(1)
Put t = 1 in (1) ⇒ A(2 – 1) + 0 = 1 ⇒ A = 1
put t = 2 in (1) ⇒ 0 + B(1 – 2) = 1 ⇒ B = -1

Question 8.
Find the integral of \(\frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)}\)
Solution:
We have \(\frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)}=\frac{x^4+3 x^2+2}{x^4+7 x^2+12}=\frac{x^4+7 x^2+12-4 x^2+10}{x^4+7 x^2+12}=1-\frac{\left(4 x^2+10\right)}{\left(x^2+3\right)\left(x^2+4\right)}\) …………(a)
Put x = t then
Let \(\frac{(4 t+10)}{(t+3)(t+4)}=\frac{A}{(t+3)}+\frac{B}{(t+4)}=\frac{A(t+4)+B(t+3)}{(t+3)(t+4)}\)
⇒ A(t + 4) + B(t + 3) = (4t + 10) ..(1)
Put t = -3 in (1) ⇒ A(-3 + 4) + B(0) = -12 + 10 = -2 ⇒ A = -2
Put t = -4 in (1) ⇒ A(0) + B(-4 + 3) = -16 + 10 = -6 ⇒ B = 6
From (1),

Question 9.
Find the integral of \(\frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)}\)
Solution:
Given integral is \(\frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)}\) Put x² = t ⇒ 2x dx = dt
∴ \(\int \frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)} d x=\int \frac{d t}{(t+1)(t+3)}\) ……(1)
Let \(\frac{1}{(t+1)(t+3)}=\frac{A}{(t+1)}+\frac{B}{(t+3)}=\frac{A(t+3)+B(t+1)}{(t+1)(t+3)}\)
⇒ A(t + 3) + B(t + 1) = 1 …..(1)
Put t = -1 in (1) ⇒ A(-1 + 3) + 0 = 1 ⇒ A = 1/2
Put t = -3 in (1) ⇒ 0 + B(-3 + 1) = 1 ⇒ B = -1/2

Question 10.
Find the integral of \(\frac{1}{x\left(x^4-1\right)}\)
Solution:
Given integrand is \(\frac{1}{x\left(x^4-1\right)}\)
Multiplying Nr and Dr by x³, we get \(\frac{1}{x\left(x^4-1\right)}=\frac{x^3}{x^4\left(x^4-1\right)}\)
∴ \(\int \frac{1}{x\left(x^4-1\right)} d x=\int \frac{x^3}{x^4\left(x^4-1\right)} d x \)
Put x = t ⇒ 4x³ = dt
∴ \(\int \frac{1}{x\left(x^4-1\right)} d x=\frac{1}{4} \int \frac{d t}{t(t-1)}\)
Let \(\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{(t-1)}=\frac{A(t-1)+B(t)}{(t-1) t}\) ⇒ A(t – 1) + Bt = 1 ………….(1)
Put t = 0 in (1) ⇒ A(0 – 1) + 0 = 1 ⇒ A = -1
Put t = 1 in(1) ⇒ 0 + B = 1 ⇒ B = 1

Question 11.
Find the integral of \(\frac{1}{x-x^3}\)
Solution:
We have \(\frac{1}{x-x^3}=\frac{1}{x\left(1-x^2\right)}=\frac{1}{x(1-x)(1+x)}\)
Let \(\frac{1}{x(1-x)(1+x)}=\frac{A}{x}+\frac{B}{(1-x)}+\frac{C}{(1+x)}=\frac{A\left(1-x^2\right)+B x(1+x)+C x(1-x)}{x(1-x)(1+x)}\)
⇒ A(1 – x²) + Bx(1 + x) + Cx(1 – x) = 1 ……….(1)
Put x = 1 in (1) ⇒ A(0) + B(1 + 1) + C(0) = 1 ⇒ 2B = 1 ⇒ B = 1/2
Put x = 0 in (1) ⇒ A(1) + B(0) + C(0) = 1 ⇒ A = 1
Comparingthe coefficient of x, B + C = 0 ⇒ C = -B = \(-\frac{1}{2}\)
A = 1, B = \(\frac{1}{2}\) , C = \(-\frac{1}{2}\)

Question 12.
Find the integral of \(\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}\) [Hint: \(\frac{1}{x^{\frac{1}{3}}\left(1+x^{\frac{1}{6}}\right)}\), put x = t⁶]
Solution:
Given that \(\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\frac{1}{x^{\frac{1}{3}}\left(1+x^{\frac{1}{6}}\right)}\)
Put x = t⁶ ⇒ dx = 6t⁵ dt

Question 13.
Find the integral of \(\frac{5 x}{(x+1)\left(x^2+9\right)}\)
Solution:
Let \(\frac{5 x}{(x+1)\left(x^2+9\right)}=\frac{A}{(x+1)}+\frac{B x+C}{\left(x^2+9\right)}=\frac{A\left(x^2+9\right)+(B x+C)(x+1)}{(x+1)\left(x^2+9\right)}\)
⇒ A(x² + 9) + (Bx + C)(x + 1) = 5x …….(1)
Put x = -1 in (1) ⇒ A(1 + 9) + (Bx + C)(0) = 5(-1) ⇒ 10A = -5 ⇒ A = -1/2
Comparingthe coefficientof x² we have B + A = 0 ⇒ B = -A = 1/2
Comparingthe coefficientof x² we have C + 9A = 0 ⇒ C = -9A = 9/2
A = \(-\frac{1}{2}\), B = \(\frac{1}{2}\), C = \(\frac{9}{2}\)

Question 14.
Find the integral of \(\frac{x^2+x+1}{(x+1)^2(x+2)}\)
Solution:
Let \(\frac{x^2+x+1}{(x+1)^2(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^2}+\frac{C}{(x+2)}=\frac{A(x+1)(x+2)+B(x+2)+C(x+1)^2}{(x+1)^2(x+2)}\)
⇒ A(x + 1)(x + 2) + B(x + 2) + C(x + 1)² = x + x + 1 ……….(1)
= A(x² + 3x + 2) + B(x + 2) + C(x² + 2x + 1)
⇒ x² + x + 1 = (A + C)x² + (3A + B + 2C)x + (2A + 2B + C)
Put x = -1 in (1) ⇒ A(0) + B(-1 + 2) + C(0) = 1 – 1 + 1 ⇒ B = 1
Put x = -2 in (1) ⇒ A(0) + B(0) + C(-2 + 1)² = (-2)² – 2 + 1 ⇒ C = 4 – 2 + 1 = 3
Equating the coefficient os x², we get A + C = 1 ⇒ A = 1 – C = 1 – 3 = -2
A = -2, B = 1, C = 3