Let $I=\int_0^1 \frac{x}{\sqrt{1+x^2}} d x$

Put $1+x^2=t^2$

$$\begin{array}{lr} \Rightarrow & 2 x d x=2 t d t \\ \Rightarrow & x d x=t d t \end{array}$$

$$\begin{aligned} \therefore\quad I & =\int_1^{\sqrt{2}} \frac{t d t}{t} \\ & =[t]_1^{\sqrt{2}}=\sqrt{2}-1 \end{aligned}$$

Let $$I=\int_0^\pi x \sin x \cos ^2 x d x\quad\text{.... (i)}$$

and $$I=\int_0^\pi(\pi-x) \sin (\pi-x) \cos ^2(\pi-x) d x$$

$\Rightarrow \quad I=\int_0^\pi(\pi-x) \sin x \cos ^2 x d x\quad\text{.... (ii)}$

On adding Eqs. (i) and (ii), we get

$$2 I=\int_0^\pi \pi \sin x \cos ^2 x d x$$

$$\begin{array}{ll} \text { Put } & \cos x=t \\ \Rightarrow & -\sin x d x=d t \end{array}$$

As $x \rightarrow 0$, then $t \rightarrow 1$

and $x \rightarrow \pi$, then $t \rightarrow-1$

$\therefore\quad I=-\pi \int_1^{-1} t^2 d t \Rightarrow I=-\pi\left[\frac{t^3}{3}\right]_1^{-1}$

$\Rightarrow\quad 2 I=-\frac{\pi}{3}[-1-1] \Rightarrow 2 I=\frac{2 \pi}{3}$

$\therefore \quad I=\frac{\pi}{3}$

Let $I=\int_0^{1 / 2} \frac{d x}{\left(1+x^2\right) \sqrt{1-x^2}}$

Put $x=\sin \theta$

$\Rightarrow \quad d x=\cos \theta d \theta$

As $\quad x \rightarrow 0$, then $\theta \rightarrow 0$

and $x \rightarrow \frac{1}{2}$, then $\theta \rightarrow \frac{\pi}{6}$

$$\begin{aligned} \therefore\quad I & =\int_0^{\pi / 6} \frac{\cos \theta}{\left(1+\sin ^2 \theta\right) \cos \theta} d \theta=\int_0^{\pi / 6} \frac{1}{1+\sin ^2 \theta} d \theta \\ & =\int_0^{\pi / 6} \frac{1}{\cos ^2 \theta\left(\sec ^2 \theta+\tan ^2 \theta\right)} d \theta \\ & =\int_0^{\pi / 6} \frac{\sec ^2 \theta}{\sec ^2 \theta+\tan ^2 \theta} d \theta \\ & =\int_0^{\pi / 6} \frac{\sec ^2 \theta}{1+\tan ^2 \theta+\tan ^2 \theta} d \theta \\ & =\int_0^{\pi / 6} \frac{\sec ^2 \theta}{1+2 \tan ^2 \theta} d \theta \end{aligned}$$

$$\begin{aligned} &\text { Again, put } \quad \tan \theta =t \\ & \Rightarrow\quad\sec ^2 \theta d \theta =d t \end{aligned}$$

As $\theta \rightarrow 0$, then $t \rightarrow 0$

and $\theta \rightarrow \frac{\pi}{6}$, then $t \rightarrow \frac{1}{\sqrt{3}}$

$$\begin{aligned} \therefore\quad I & =\int_0^{1 / \sqrt{3}} \frac{d t}{1+2 t^2}=\frac{1}{2} \int_0^{1 / \sqrt{3}} \frac{d t}{\left(\frac{1}{\sqrt{2}}\right)^2+t^2} \\ & =\frac{1}{2} \cdot \frac{1}{1 / \sqrt{2}}\left[\tan ^{-1} \frac{t}{\frac{1}{\sqrt{2}}}\right]_0^{1 / \sqrt{3}}=\frac{1}{\sqrt{2}}\left[\tan ^{-1}(\sqrt{2} t)\right]_0^{1 / \sqrt{3}} \\ & =\frac{1}{\sqrt{2}}\left[\tan ^{-1} \sqrt{\frac{2}{3}}-0\right]=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\sqrt{\frac{2}{3}}\right) \end{aligned}$$

Let $I=\int \frac{x^2}{x^4-x^2-12} d x$

$$\begin{aligned} & =\int \frac{x^2}{x^4-4 x^2+3 x^2-12} d x \\ & =\int \frac{x^2 d x}{x^2\left(x^2-4\right)+3\left(x^2-4\right)} \\ & =\int \frac{x^2 d x}{\left(x^2-4\right)\left(x^2+3\right)} \\ \text{Now,}\quad & \frac{x^2}{\left(x^2-4\right)\left(x^2+3\right)}\quad \left[\text { let } x^2=t\right] \end{aligned}$$

$\Rightarrow \quad \frac{t}{(t-4)(t+3)}=\frac{A}{t-4}+\frac{B}{t+3}$

$\Rightarrow \quad t=A(t+3)+B(t-4)$

On comparing the coefficient of t on both sides, we get

$$\begin{array}{lrl} & A+B=1 \quad\text{.... (i)}\\ \Rightarrow & 3 A-4 B=0 \quad\text{.... (ii)}\\ \Rightarrow & 3(1-B)-4 B=0 \\ \Rightarrow & 3-3 B-4 B=0 \\ \Rightarrow & 7 B=3 \\ \Rightarrow & B=\frac{3}{7} \end{array}$$

If $B=\frac{3}{7}$, then $A+\frac{3}{7}=1$

$\Rightarrow\quad A=1-\frac{3}{7}=\frac{4}{7}$

$\frac{x^2}{\left(x^2-4\right)\left(x^2+3\right)}=\frac{4}{7\left(x^2-4\right)}+\frac{3}{7\left(x^2+3\right)}$

$$\begin{aligned} \therefore\quad I & =\frac{4}{7} \int \frac{1}{x^2-(2)^2} d x+\frac{3}{7} \int \frac{1}{x^2+(\sqrt{3})^2} d x \\ & =\frac{4}{7} \cdot \frac{1}{2 \cdot 2} \log \left|\frac{x-2}{x+2}\right|+\frac{3}{7} \cdot \frac{1}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}+C \\ & =\frac{1}{7} \log \left|\frac{x-2}{x+2}\right|+\frac{\sqrt{3}}{7} \tan ^{-1} \frac{x}{\sqrt{3}}+C \end{aligned}$$

Let $I=\int \frac{x^2}{\left(x^2+a^2\right)\left(x^2+b^2\right)} d x$

Now, $\frac{x^2}{\left(x^2+a^2\right)\left(x^2+b^2\right)}\quad \left[\text { let } x^2=t\right]$

$$\begin{aligned} & =\frac{t}{\left(t+a^2\right)\left(t+b^2\right)}=\frac{A}{\left(t+a^2\right)}+\frac{B}{\left(t+b^2\right)} \\ & t=A\left(t+b^2\right)+B\left(t+a^2\right) \end{aligned}$$

On comparing the coefficient of t, we get

$$\begin{aligned} A+B & =1 \quad\text{.... (i)}\\ b^2 A+a^2 B & =0 \quad\text{.... (i)}\\ \Rightarrow\quad b^2(1-B)+a^2 B & =0 \\ \Rightarrow\quad b^2-b^2 B+a^2 B & =0 \\ \Rightarrow\quad b^2+\left(a^2-b^2\right) B & =0 \\ \Rightarrow\quad B=\frac{-b^2}{a^2-b^2} & =\frac{b^2}{b^2-a^2} \end{aligned}$$

From Eq. (i), $A+\frac{b^2}{b^2-a^2}=1$

$\Rightarrow\quad A=\frac{b^2-a^2-b^2}{b^2-a^2}=\frac{-a^2}{b^2-a^2}$

$$\begin{aligned} \therefore\quad I & =\int \frac{-a^2}{\left(b^2-a^2\right)\left(x^2+a^2\right)} d x+\int \frac{b^2}{b^2-a^2} \cdot \frac{1}{x^2+b^2} d x \\ & =\frac{-a^2}{\left(b^2-a^2\right)} \int \frac{1}{x^2+a^2} d x+\frac{b^2}{b^2-a^2} \int \frac{1}{x^2+b^2} d x \\ & =\frac{-a^2}{b^2-a^2} \cdot \frac{1}{a} \tan ^{-1} \frac{x}{a}+\frac{b^2}{b^2-a^2} \cdot \frac{1}{b} \tan ^{-1} \frac{x}{b} \\ & =\frac{1}{b^2-a^2}\left[-\tan ^{-1} \frac{x}{a}+b \tan ^{-1} \frac{x}{b}\right] \\ & =\frac{1}{a^2-b^2}\left[\tan ^{-1} \frac{x}{a}-b \tan ^{-1} \frac{x}{b}\right] \end{aligned}$$