e ^ 3 x (x^4+1 )^ -1 d x - Vidyadip

Question

$\int e ^{3 \log x}\left(x^4+1\right)^{-1} d x$

✓

Answer

$ \text { Let } I =\int e ^{3 \log x}\left(x^4+1\right)^{-1} d x$
$=\int \frac{ e ^{\log \left(x^3\right)}}{x^4+1} d x$
$=\int \frac{x^3}{x^4+1} d x $
Put $x^4+1=t$
Differentiating w.r.t. $x$, we get
$ 4 x ^3 dx = dt$
$\therefore x ^3 dx =\frac{1}{4} dt$
$\therefore I =\frac{1}{4} \int \frac{ dt }{ t }$
$=\frac{1}{4} \log | t |+ c$
$\therefore I =\frac{1}{4} \log \left|x^4+1\right|+ 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

Explore more

More "Solve the Following Question.(2 Marks)" from Question Bank [2022]→Question Bank [2022] — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

If $\text{f}'\text{(x)}=\text{x}-\frac{1}{\text{x}^2}$ and $\text{f}'(1)=\frac{1}{2}$, find f'(x).

Prove The Following : $\int \frac{1}{a^2-x^2} \cdot d x=\frac{1}{2 a} \log \left(\frac{a+x}{a-x}\right)+c$

Evaluate the following integrals:
$\int(\sec^2\text{x}+\text{cosec}^2\text{x})\text{dx}$

Compute the products AB and BA whichever exists the following cases:
$\text{A}=\begin{bmatrix}3&2\\-1&0\\-1&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}4&5&6\\0&1&2\end{bmatrix}$

Using the rules in logic, write the negations of the following : (p → q) ∧ r

Construct a $2 \times 2$ matrix $A = [a_{ij}]$ whose elements $a_{ij}$ are given by:
$\text{a}_\text{ij}=\frac{(\text{i}-\text{j})^2}{2}$

In $\triangle ABC,$ if $a = 13, b = 14, c = 15,$ then find the value of $\cos B$

Differentiate the following w. r. t. x.$\sin ^{-1}\left(\frac{a \cos x-b \sin x}{\sqrt{a^2+b^2}}\right)$

Evaluate the following : $\int\left(\frac{\cos x}{1-\cos x}\right) \cdot d x$

"Let $f(x)=x^2+5$ and $g(x)=e^x+3$ then
$f[g(x)]=$ and $g[f(x)]=$
Now $f^{\prime}(x)=$ and $g^{\prime}(x)=$
The derivative off $[g(x)]$ w.r. t. $x$ in terms of $f$ and $g$ is
Therefore $\frac{d}{d x}[f[g(x)]]=$ and $\left[\frac{d}{d x}[f[g(x)]]\right]_x=0=$
The derivative of $g[f(x)]$ w.r. t. $x$ in terms of $f$ and $g$ is
Therefore $\frac{d}{d x}[ g [ f ( x )]]=\ldots . . . .$. and $\left[\frac{d}{d x}[ g [ f ( x )]]\right]_{ x =1}=$
Hint basket: $\left\{f^{\prime}[g(x)] \cdot g^{\prime}(x), 2 e^{2 x}+6 e^x, 8, g^{\prime}[f(x)] \cdot f^{\prime}(x), 2 x e^{x 2+5},-2 e^6, e^{2 x}+6 e^x+14, e^{x 2}+\right.$ $\left.5+3,2 x, e^x\right\}$