Download App

Indefinite Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIndefinite Integrals5 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 Indefinite Integrals→Indefinite Integrals — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Find $\frac{\text{dy}}{\text{dx}},\text{if y = sin}^{-1}[\text{x}\sqrt{1 - x}-\sqrt{x}\sqrt{1 - x^{2}}].$
→
Find the area bounded by the curve $\text{y}=\cos\text{x}$, x-axis and the ordinates $\text{x}=0, \text{x}=2\pi$.
→
Solve the following systems of linear equations by cramer's rule:
2x - y = -2,
3x + 4y = 3
→
Prove that: $\begin{vmatrix}\text{a}^2&\text{a}^2-(\text{b}-\text{c})^2&\text{bc}\\\text{b}^2&\text{b}^2-(\text{c}-\text{a})^2&\text{ca}\\\text{c}^2&\text{c}^2-(\text{a}-\text{b})^2&\text{ab}\end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2)$
→
Find one-parameter families of solution curves of the following differential equation: (or solve the following differential equation)$\text{x}\log\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=2\log\text{x}$
→
The two adjacent sides of a parallelogram are $2\hat{i} - 4\hat{j} - 5\hat{k} \text{ and } 2\hat{i} + 2\hat{j} + 3\hat{k}.$ Find the two unit vectors parallel its diagonals. Using the diagonal vectors, find the area of the parallelogram.
→
Evaluate:
$\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
→
show that the differential equation of which $\text{y}=2(\text{x}^2-1)+\text{ce}^{-\text{x}^2}$ is a solution of the differential equation $\frac{\text{dy}}{\text{dx}}+2\text{xy}=4\text{x}^3$
→
Solve the following differential equation:
$\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}-\text{x}\cos^2\Big(\frac{\text{y}}{\text{x}}\Big)$
→
Evaluate:$\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0}\frac{\text{tan x - sin x}}{\sin^{3}\text{x}}$.
→