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

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

Answers (1)

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}

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