The value of $\int_{-1}^1 x|x| d x$ is

$$\int_\limits{-1}^1 \frac{|x-2|}{x-2} d x, x \neq 2 \text { is equal to: }$$

Assertion (A): $$\int_\limits2^8 \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x=3$$

Reason (R): $$\int_\limits a^b f(x) d x=\int_a^b f(a+b-x) d x$$

(a) Evaluate: $\int_0^\pi \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x$

OR

(b) Find: $\int \frac{2 x+1}{(x+1)^2(x-1)} d x$

(a) Evaluate: $$\int_\limits0^{2 \pi} \frac{1}{1+e^{\sin x}} d x$$

OR

$$\text { (b) Find: } \int \frac{x^4}{(x-1)\left(x^2+1\right)} d x$$

Evaluate : $$\int_\limits0^{\frac{\pi}{2}} \sin 2 x \tan ^{-1}(\sin x) d x$$