Download App

Indefinite Integrals — MATHS STD 12 Science — Question

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

Answer

Consider the integrals
$\text{I}=\int\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}\ \text{dx}$
Now let us separate the fraction $\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}$
Through particle fractions.
$\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$
$\Rightarrow\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}=\frac{\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(\text {x}-1)}{(\text{x}^2+1)(\text{x}-1)}$
$\Rightarrow\text{x}=\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(\text{x}-1)$
Compairing the coefficient, we have,
A + B = 0, -B + C = 1 and A - C = 0
Solving the equations we, get,
$\Rightarrow\text{A}=\frac{1}{2},\text{B}=-\frac{1}{2}\text{ and }\text{C}=\frac{1}{2}$
$\Rightarrow\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$
$\Rightarrow\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}=\frac{1}{2}\times\frac{1}{\text{x}-1}-\frac{1}{2}\times\frac{\text{x}-1}{\text{X}^2+1}$
$\Rightarrow\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}=\frac{1}{2(\text{x}-1)}-\frac{\text{x}}{2(\text{x}^2+1)}+\frac{1}{2(\text{x}^2+1)}$
Thus, we have,
$=\int\frac{\text{x}}{(\text{x}^2+1)(\text{x}-1)}\ \text{dx}$
$=\int\Big[\frac{1}{2(\text{x}-1)}-\frac{\text{x}}{2(\text{x}^2+1)}+\frac{1}{2(\text{x}^2+1)}\Big]\text{dx}$
$=\int\frac{\text{dx}}{2(\text{x}-1)}-\int\frac{\text{xdx}}{2(\text{x}^2+1)}+\int\frac{\text{dx}}{2(\text{x}^2+1)}$
$=\int\frac{1}{2}\frac{\text{dx}}{(\text{x}-1)}-\int\frac{1}{2}\frac{\text{xdx}}{(\text{x}^2+1)}+\int\frac{1}{2}\frac{\text{dx}}{(\text{x}^2+1)}$
$=\frac{1}{2}\int\frac{\text{dx}}{(\text{x}-1)}-\frac{1}{2}\times\frac{1}{2}\int\frac{2\text{xdx}}{(\text{x}^2+1)}+\frac{1}{2}\int\frac{\text{dx}}{(\text{x}^2+1)}$
$=\frac{1}{2}\log|\text{x}-1|-\frac{1}{4}\log(\text{x}^2+1)+\frac{1}{2}\tan^{-1}\text{x}+\text{C}$

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

Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\text{x}+2\text{y})}{\text{x}(2\text{x}+\text{y})},\text{y}(1)=2$
→
Find the area of the region $\Bigg\{(\text{x},\text{y}): \frac{\text{x}^{2}}{\text{a}^{2}}+\frac{\text{y}^{2}}{\text{b}^{2}}<1< \frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}\bigg\}$
→
Evaluate the following integrals:
$\int(\text{x}-3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
→
Integrate the function in Exercise:$\frac{1}{\sqrt{\sin^{3}\text{x}\sin(\text{x}+\text{a)}}}$
→
Prove the following identities: $\begin{vmatrix}\text{y}+\text{z}&\text{z}&\text{y}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}=4\text{xyz}$
→
Maximize $Z = -x_1 + 2x_2$ Subject to $-\text{x}_1+3\text{x}_2\leq10 , \text{x}_1+\text{x}_2\leq6 , \text{x}_1+\text{x}_2\leq2 , \text{x}_1 \text{x}_2\geq0$
→
Solve the differential equation:
$(\tan^{-1}\text{y} - x) \text{dy} = ( 1 + \text{y}^{2}) \text{dx}$
→
Using differentials, find the approximate values of the following:
$\sqrt{26}$
→
Solve the following differential equation:
$\text{dx + xdy}=\text{e}^{-\text{y}}\sec^2\text{y dy}$
→
Differentiate $\sin^{-1}\Big\{\frac{2^{\text{x}+1}\times3^\text{x}}{1+(36)^\text{x}}\Big\}$ with respect to x:
→