Question
$\int \frac{\left(x^2+2\right)}{x^2+1} a ^{x+\tan ^{-1 x}} d x$

Answer

Let $I =\int\left(\frac{x^2+2}{x^2+1}\right) a ^{x+\tan ^{-1 x}} d x$
Put $x+\tan ^{-1} x=t$
Differentiating w.r.t. $x$, we get
$ \left(1+\frac{1}{1+x^2}\right) d x= dt$
$\therefore\left(\frac{x^2+2}{x^2+1}\right) d x= dt$
$\therefore I =\int a ^1 dt$
$=\frac{ a ^1}{\log a }+ c$
$\therefore I =\frac{ a ^{x+\tan ^{-1 x}}}{\log a }+ c $

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Evaluate the following definite integrals:$\int_{0}^\limits{\pi}\Big(\sin^2\frac{\text{x}}{2}-\cos^2\frac{\text{x}}{2}\Big)\text{dx}$
If $\text{y}=3\text{e}^{2\text{x}}+2\text{e}^{3\text{x}}$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}-5\frac{\text{dy}}{\text{dx}}+6\text{y}=0$
If $\vec{\text{a}},\vec{\text{b}}$ are two vectors, then write the truth value of the following statement:$\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|\Rightarrow\vec{\text{a}}=\pm\vec{\text{b}}$
Find the points o local maxima or local minima, if any, of the following functions, using the first derivatives test. Also, find the local maximum or local minimum values, as the case may be:
$\text{f}(\text{x})=\cos{\text{x}},0\leq\text{x}\leq\pi$
If the position vectors of the points A(3, 4), B(5, -6) and C(4, -1) are $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ respectively, compute $\vec{\text{a}}+2\vec{\text{b}}-3\vec{\text{c}}$.
Evaluate: $\int_0^{\frac{\pi}{2}} \frac{\cos x}{(1+\sin x)(2+\sin x)} d x$
If $\text{x}=\text{a}(\cos2\text{t}+2\text{t}\sin2\text{t})\ \text{and}\ \text{y}=\text{a}(\sin2\text{t}-2\text{t}\cos2\text{t}),$ then find $\frac{\text{d}^2\text{y}}{\text{dx}^2}.$
Find the area of the region bounded by the following curve, the X-axis and the given lines:
(i) 0 ≤ x ≤ 5, 0 ≤ y ≤ 2
(ii) y = sin x, x = 0, x = π
(iii) y = sin x, x = 0, x = $\frac{\pi}{3}$
Find the distance between the parallel lines $\frac{x}{2}=\frac{y}{-1}=\frac{z}{2}$ and $\frac{x-1}{2}=\frac{y-1}{-1}=\frac{z-3}{2}$
Prove the following results:
$2\tan^{-1}\frac{3}{4}-\tan^{-1}\frac{17}{31}=\frac{\pi}{4}$