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#### Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 10 maths textbook solution.

Answer:- $\frac{\pi}{4}$

Hints:-  You must know about the integration rules of functions .

Given:-  $\int_{0}^{\infty} \frac{x}{(1+x)\left(1+x^{2}\right)} d x$

Solution : $\int_{0}^{\infty} \frac{x}{(1+x)\left(1+x^{2}\right)} d x$

We can write integrand using partial fractions

\begin{aligned} &\frac{x}{\left(x^{2}+1\right)(x+1)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C}{(x+1)} \\ &\frac{x}{\left(x^{2}+1\right)(x+1)}=\frac{(A x+B)(x+1)+C\left(x^{2}+1\right)}{\left(x^{2}+1\right)(x+1)} \end{aligned}

Cancelling denominators

$x=(A x+B)(x+1)+C\left(x^{2}+1\right)$                            ......(1)

Putting $x=1, \text { in (1) }$

\begin{aligned} &x=(A x+B)(x+1)+C\left(x^{2}+1\right) \\ &1=(A \times 1+B)(1+1)+C\left(1^{2}+1\right) \\ &1=2(A+B)+C(2) \\ &1=2 A+2 B+2 C \end{aligned}                  .....(2)

Putting $x=0 \text { in (1) }$

\begin{aligned} &x=(A x+B)(x+1)+C\left(x^{2}+1\right) \\ &0=(A \times 0+B)(0+1)+C(0+1) \end{aligned}

\begin{aligned} &0=B+C \\ &B=-C \end{aligned}

Put in (2)

\begin{aligned} &1=2 A-2 C+2 C \\ &A=\frac{1}{2} \end{aligned}

Putting $x = -1$

\begin{aligned} &-1=\left(A(-1)+B(-1+1)+C\left((-1)^{2}+1\right)\right. \\ &-1=(-A+B)(0)+C(2) \\ &-1=2 C \\ &C=\frac{-1}{2} \end{aligned}

Put value of A and C in (2)

\begin{aligned} &2 A+2 B+2 C=1 \\ &2 \times \frac{1}{2}+2 B+2 \times \frac{-1}{2}=1 \\ &1+2 B-1=1 \\ &2 B=1 \\ &B=\frac{1}{2} \end{aligned}

$\therefore A=\frac{1}{2}, B=\frac{1}{2}, C=\frac{-1}{2}$

Hence we can write

$\begin{gathered} \frac{x}{(x+1)\left(x^{2}+1\right)}=\frac{\left(\frac{1}{2} x+\frac{1}{2}\right)}{\left(x^{2}+1\right)}+\frac{\frac{-1}{2}}{(x+1)} \\ =\frac{x}{2\left(x^{2}+1\right)}+\frac{1}{2\left(x^{2}+1\right)}-\frac{1}{2(x+1)} \end{gathered}$

Solving

\begin{aligned} &I_{3}=\frac{-1}{2} \int_{0}^{\infty} \frac{1}{x+1} d x \\ &=\frac{-1}{2}\left[\log |x+1|_{0}^{\infty}\right. \\ &=\frac{-1}{2}[\log (1)-\log (1) \mid] \\ &=0 \end{aligned}

Hence,

\begin{aligned} &I=I_{1}+I_{2}+I_{3} \\ &=0+\frac{\pi}{4}+0 \\ &=\frac{\pi}{4} \end{aligned}