#### Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 17 maths textbook solution

$\frac{\pi }{4}$

Given:

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

Hint:

You must know derivation of sinx, cosx and tanx.

Explanation:

Let

\begin{aligned} &I=\int_{0}^{\infty} \frac{x}{(1+x)\left(1+x^{2}\right)} d x \\\\ &=\int_{0}^{\frac{\pi}{2}} \frac{\tan \theta}{(1+\tan \theta)\left(1+\tan ^{2} \theta\right)} \cdot \sec ^{2} \theta d \theta \end{aligned}

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\tan \theta}{(1+\tan \theta)\left(\sec ^{2} \theta\right)} \cdot \sec ^{2} \theta d \theta \\\\ &=\int_{0}^{\frac{\pi}{2}} \frac{\tan \theta+1-1}{1+\tan \theta} d \theta \end{aligned}

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} 1 d \theta-\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan \theta} d \theta \\\\ &=\int_{0}^{\frac{\pi}{2}} 1 d \theta-\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\frac{\sin \theta}{\cos \theta}} d \theta, \quad\left[\therefore \tan \theta=\frac{\sin \theta}{\cos \theta}\right] \end{aligned}

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} 1 d \theta-\int_{0}^{\frac{\pi}{2}} \frac{\cos \theta}{\cos \theta+\sin \theta} d \theta \\\\\ &=[\theta]_{0}^{\frac{\pi}{2}}-I_{1}, \quad \text { where } I_{1}=\int_{0}^{\frac{\pi}{2}} \frac{\cos \theta}{\cos \theta+\sin \theta} d \theta \end{aligned}

\begin{aligned} &=\left(\frac{\pi}{2}-0\right)-I_{1} \\\\ &=\frac{\pi}{2}-I_{1} \ldots(i) \end{aligned}

Now,

\begin{aligned} &I_{1}=\int_{0}^{\frac{\pi}{2}} \frac{\cos \theta}{\cos \theta+\sin \theta} d \theta \ldots(i i) \\\\ &=\int_{0}^{\frac{\pi}{2}} \frac{\cos \left(\frac{\pi}{2}-\theta\right)}{\cos \left(\frac{\pi}{2}-\theta\right)+\sin \left(\frac{\pi}{2}-\theta\right)} d \theta \end{aligned}

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\sin \theta}{\sin \theta+\cos \theta} d \theta, \quad\left[\therefore \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right] \\\\ &I_{1}=\int_{0}^{\frac{\pi}{2}} \frac{\sin \theta}{\sin \theta+\cos \theta} d \theta \ldots(i i i) \end{aligned}

\begin{aligned} &I_{1}=\int_{0}^{\frac{\pi}{2}} \frac{\cos \theta+\sin \theta}{\sin \theta+\cos \theta} d \theta \\\\ &=\int_{0}^{\frac{\pi}{2}} 1 d \theta \end{aligned}

\begin{aligned} &=\frac{1}{2}[\theta]_{0}^{\frac{\pi}{2}} \\\\ &=\frac{1}{2}\left[\frac{\pi}{2}-0\right] \\\\ &=\frac{\pi}{4} \end{aligned}

Put in (i)

\begin{aligned} &I=\frac{\pi}{2}-\frac{\pi}{4} \\\\ &I=\frac{2 \pi-\pi}{4} \\\\ &I=\frac{\pi}{4} \end{aligned}