Download App

Indefinite Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIndefinite Integrals5 Marks
Question
Evaluate the following integrals:
$\int\frac{1}{(\text{x}^2+2\text{x}+10)^2}\text{ dx}$
✓

Answer

Let $\int\frac{1}{(\text{x}^2+2\text{x}+10)^2}\text{ dx}$
$=\int\frac{1}{\big[(\text{x}+1)^2+3^2\big]}\text{ dx}$
Let $\text{x}+1=3\tan\theta$
On differentiating both sides, we get
$\text{dx}=3\sec^2\theta\text{ d}\theta$
$\therefore\ \text{I}=\int\frac{1}{\big[3^2\tan^2\theta+3^2\big]^2}3\sec^2\theta\text{ d}\theta$
$=\frac{1}{27}\int\frac{\sec^2\theta}{\sec^4\theta}\text{ d}\theta$
$=\frac{1}{27}\int\frac{1}{\sec^2\theta}\text{ d}\theta$
$=\frac{1}{27}\int\cos^2\theta\text{ d}\theta$
$=\frac{1}{54}\int(1+\cos2\theta)\text{d}\theta$
$=\frac{1}{54}\Big(\theta+\frac{\sin2\theta}{2}\Big)+\text{ C}$
$=\frac{1}{54}\Big(\theta+\frac{\tan\theta}{1+\tan^2\theta}\Big)+\text{C}$
$=\frac{1}{54}\begin{pmatrix}\tan^{-1}\frac{\text{x}+1}{3}+\frac{\tan\Big(\tan^{-1}\frac{\text{x}+1}{3}\Big)}{1+\tan^{2}\Big(\tan^{-1}\frac{\text{x}+1}{3}\Big)}\end{pmatrix}+\text{C}$
$=\frac{1}{54}\begin{pmatrix}\tan^{-1}\frac{\text{x}+1}{3}+\frac{\frac{\text{x}+1}{3}}{1+\Big(\frac{\text{x}+1}{3}\Big)^2}\end{pmatrix}+\text{C}$
$=\frac{1}{54}\Bigg(\tan^{-1}\frac{\text{x}+1}{3}+\frac{\frac{\text{x}+1}{3}}{\frac{\text{x}^2+2\text{x}+10}{9}}\Bigg)+\text{C}$
$=\frac{1}{54}\bigg(\tan^{-1}\frac{\text{x}+1}{3}+\frac{3(\text{x}+1)}{\text{x}^2+2\text{x}+10}\bigg)+\text{C}$

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 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

Evaluvate the following intregals:
$\int\frac{1}{1-\tan\text{x}}\text{ dx}$
→
Evaluate the following integrals:
$\int\frac{\text{x}^2+1}{\text{x}^4+\text{x}^2+1}\ \text{dx}$
→
Maximum Z = 4x + 3y
Subject to
$3\text{x}+4\text{y}\leq24$
$8\text{x}+6\text{y}\leq48$
$\text{x}\leq5$
$\text{y}\leq6$
$\text{x},\text{y}\geq0$
→
Using elementary row operations, find the inverse of the following matrix:$\begin{pmatrix} 2 & 5 \\ 1 & 3 \\ \end{pmatrix}.$
→
$\text{If y}=\text{e}^{\text{a}\cos^{-1}\text{x}},-1\leq\text{x}\leq1,\ \text{show that}(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}-\text{a}^2\text{y}=0.$
→
Find the intervals in which the following functions are increasing or decreasing. $f(x) = x^{2 }+ 2x - 5$
→
Solve the system of equations
$ \frac{2}{x} + \frac{3}{y} + \frac{{10}}{z} = 4$
$ \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1$
$ \frac{6}{x} + \frac{9}{y} - \frac{{ 20}}{z} = 2$
→
Differentiate the following functions with respect to x:
$\text{x}\sin2\text{x}+5^{\text{x}}+\text{k}^\text{k}+(\tan^2\text{x})^3$
→
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point $'c'$ in the indicated interval as stated by the Lagrange's mean value theorem.
$f(x) = x^{3 }- 5x^2 - 3x$ on $[1, 3]$
→
A and B are partners sharing profits and losses in the ratio 3 : 4 respectively. They admit C as a new partner, the new profit sharing ratio being 2 : 2 : 3 between A, B and C respectively. C pays Rs. 12,000 as premium for goodwill. Find the amount of premium shared by A and B.
→