Download App

INTEGRALS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS5 Marks
Question
Evaluate the following integrals:$\int\frac{\text{x}^3}{\text{x}^4+\text{x}^2+1}\text{ dx}$
✓

Answer

$\text{I}=\int\frac{\text{x}^3}{\text{x}^4+\text{x}^2+1}\text{ dx}$ $=\int\frac{\text{x}^2\cdot\text{x}}{(\text{x}^2)^2+\text{x}^2+1}\text{ dx}$ Let $\text{x}^2=\text{t}$ or $2\text{x}\text{ dx}=\text{dt}$ $\text{I}=\frac{1}{2}\int\frac{\text{t}}{\text{t}^2+\text{t}+1}\text{ dt}$ $=\frac{1}{4}\int\frac{2\text{t}}{\text{t}^2+\text{t}+1}\text{ dt}$$=\frac{1}{4}\int\frac{2\text{t}+1-1}{\text{t}^2+\text{t}+1}\text{ dt}$
$=\frac{1}{4}\int\Big[\frac{(2\text{t}+1)}{(\text{t}^2+\text{t}+1)}-\frac{1}{(\text{t}^2+\text{t}+1)}\Big]\text{dt}$
$=\frac{1}{4}\Big[\log\big|\text{t}^2+\text{t}+1\big|-\int\frac{1}{\big(\text{t}^2+\text{t}+\frac{1}{4}+\frac{3}{4}\big)}\text{ dt}\Big]$ $=\frac{1}{4}\begin{bmatrix}\log\big|\text{t}^2+\text{t}+1\big|-\int\frac{1}{\big(\text{t}+\frac{1}{2}\big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}\text{ dt}\end{bmatrix}$ $=\frac{1}{4}\begin{bmatrix}\log\big|\text{t}^2+\text{t}+1\big|-\frac{2}{\sqrt3}\tan\frac{\big(\text{t}+\frac{1}{2}\big)}{\Big(\frac{\sqrt3}{2}\Big)}\end{bmatrix}+\text{C}$ $=\frac{1}{4}\Big[\log\big|\text{t}^2+\text{t}+1\big|-\frac{2}{\sqrt3}\tan\Big(\frac{2\text{t}+1}{\sqrt3}\Big)\Big]+\text{C}$ $=\frac{1}{4}\Big[\log\big|\text{x}^4+\text{x}^2+1\big|-\frac{2}{\sqrt3}\tan\Big(\frac{2\text{x}^2+1}{\sqrt3}\Big)\Big]+\text{C}$

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

The equation of tangent at (2, 3) on the curve $y^{2} = \text{ax}^{3} + \text{b is y = 4x - 5}. $ Find the value of a and b.
→
Find the shortest distance between the lines
$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$ and, $\vec{\text{r}}=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}+\mu\big(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$
→
$\begin{bmatrix}2&3\\5&7\end{bmatrix}\begin{bmatrix}1&-3\\-2&4\end{bmatrix}=\begin{bmatrix}-4&6\\-9&\text{x}\end{bmatrix}$ find x.
→
Differentiate $\tan^{-1}\Big(\frac{\cos\text{x}}{1+\sin\text{x}}\Big)$ with respect to $\sec^{-1}\text{x}$
→
Write the following in the simplest form:
$\tan^{-1}\sqrt{\frac{\text{a}-\text{x}}{\text{a}+\text{x}}},-\text{a}<\text{x}<\text{a}$
→
A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is $10 m$. Find the dimensions of the window to admit maximum light through the whole opening
→
Solve the following determinant equations:
$\begin{vmatrix}\text{x}+\text{a}&\text{x}&\text{x}\\\text{x}&\text{x}+\text{a}&\text{x}\\\text{x}&\text{x}&\text{x}+\text{a}\end{vmatrix}=0,\text{a}\neq0$
→
Solve the following differential equation:
$\text{y}^2\text{dx}+(\text{x}^2-\text{xy}+\text{y}^2)\text{dy}=0$
→
Maximize Z = 18x + 10y
Subject to
$4\text{x}+\text{y}\geq20$
$2\text{x}+3\text{y}\geq30$
$\text{x},\text{y}\geq0$
→
Find the equation of the plane through the points $(3, 4, 1)$ and $(0, 1, 0)$ and parallel to the line $\frac{\text{x}+3}{2}=\frac{\text{y}-3}{7}=\frac{\text{z}-2}{5}.$
→