Download App

INTEGRALS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS5 Marks
Question
Integrate the function in Exercise: $\frac{\sqrt{\text{x}^{2}+1}\big[\log\text{(x}^{2}+1)-2\log\text{x}\big]}{\text{x}^{4}}$ 
✓

Answer

$\frac{\sqrt{\text{x}^{2}+1}\big[\log\text{(x}^{2}+1)-2\log\text{x}\big]}{\text{x}^{4}}=\frac{\sqrt{\text{x}^{2}+1}}{\text{x}^{4}}\big[\log\text{(x}^{2}+1)-\log\text{x}^{2}\big]$
$=\frac{\sqrt{\text{x}^{2}+1}}{\text{x}^{4}}\bigg[\log\bigg(\frac{\text{x}^{2}+1}{\text{x}^{2}}\bigg)\bigg]$
$=\frac{\sqrt{\text{x}^{2}+1}}{\text{x}^{4}}\log\bigg(1+\frac{1}{\text{x}^{2}}\bigg)$
$=\frac{1}{\text{x}^{3}}\sqrt{1+\frac{\text{x}^{2}+1}{\text{x}^{2}}}\log\bigg(1+\frac{1}{\text{x}^{2}}\bigg)$
$=\frac{1}{\text{x}^{3}}\sqrt{1+\frac{1}{\text{x}^{2}}}\log\bigg(1+\frac{1}{\text{x}^{2}}\bigg)$
$\text{Let}\ 1+\frac{1}{\text{x}^{2}}=\text{t}\Rightarrow\frac{-2}{\text{x}^{3}}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{1}{\text{x}^{3}}\sqrt{1+\frac{1}{\text{x}^{2}}}\log\bigg(1+\frac{1}{\text{x}^{2}}\bigg)\text{dx}$
$=-\frac{1}{2}\int\sqrt{\text{t}}\log\text{t dt}$
$=-\frac{1}{2}\int\text{t}^{\frac{1}{2}}.\log\text{t dt}$
Integrating by parts, we obtain
$\text{I}=-\frac{1}{2}\Bigg[\log\text{t}.\int\text{t}^{\frac{1}{2}}\text{dt}-\left\{\bigg(\frac{\text{d}}{\text{dt}}\log\text{t}\bigg)\int\text{t}^{\frac{1}{2}}\text{dt}\right\}\text{dt}\Bigg]$
$=-\frac{1}{2}\begin{bmatrix}\log\text{t}.\frac{\text{t}\frac{3}{2}}{\frac{3}{2}}-\int\frac{1}{\text{t}}.\frac{\text{t}\frac{3}{2}}{\frac{3}{2}}\text{dt} \\ \end{bmatrix}$
$=-\frac{1}{2}\Bigg[\frac{2}{3}\text{t}^{\frac{3}{2}}\log\text{t}-\frac{2}{3}\int\text{t}^{\frac{1}{2}}\text{dt}\Bigg]$
$=-\frac{1}{2}\Bigg[\frac{2}{3}\text{t}^{\frac{3}{2}}\log\text{t}-\frac{4}{9}\int\text{t}^{\frac{3}{2}}\text{dt}\Bigg]$
$=-\frac{1}{3}\text{t}^{\frac{3}{2}}\log\text{t}+\frac{2}{9}\text{t}^{\frac{3}{2}}$
$=-\frac{1}{3}\text{t}^{\frac{3}{2}}\bigg[\log\text{t}-\frac{2}{3}\bigg]$
$=-\frac{1}{3}\bigg(1+\frac{1}{\text{x}^{2}}\bigg)^{\frac{3}{2}}\Bigg[\log\bigg(1+\frac{1}{\text{x}^{2}}\bigg)-\frac{2}{3}\Bigg]+\text{C}$

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 "5 Marks Questions" from INTEGRALS→INTEGRALS — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

$\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are the position vectors of points A, B and C respectively, prove that:
$\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{a}}$ is a vector perpendicular to the plane of triangle ABC.
→
Find which of the function:
$\text{f(x)}=\begin{cases}|\text{x}|\cos\frac{1}{\text{x}},&\text{if x}\neq0\\0,&\text{if x}=0\end{cases}$
at x = 0
→
Solve the following differential equations : $\sqrt{1+\text{x}^2+\text{y}^2+\text{x}^2\text{y}^2}+\text{xy}\ \frac{\text{dy}}{\text{dx}}=0$
→
Solve the differential equation:
$y + x \frac{dy}{dx} = x - y \frac{dy}{dx}$
→
Find the equation of the plane passing through the points (-1, 2, 0), (2, 2, -1) and parallel to the line $\frac{\text{x}-1}{1}=\frac{2\text{y}+1}{2}=\frac{\text{z}+1}{-1}.$ 
→
If $\text{u}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ and $\text{v}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big),$ where $-1 < x < 1$ then write the value of $\frac{\text{dy}}{\text{dx}}.$
→
Evaluate the following integrals:
$\int\limits_{0}^{\text{a}}\frac{\text{x}}{\sqrt{\text{a}^2+\text{x}^2}}\text{ dx}$
→
Evaluate the following integrals:$\int\frac{2\text{x}+5}{\text{x}^2-\text{x}-2}\text{ dx}$
→
$\int\text{e}^{-3\text{x}}\cos^3\text{x dx}$
→
Solve the following initial value problems:
$(\text{y}^4-2\text{x}^3\text{y})\text{dx}+(\text{x}^4-2\text{xy}^3)\text{dy}=0,\text{y}(1)=1$
→