Download App

Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
Question
Integrate the function $\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}$
✓

Answer

Given: $\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}$ 
Let $I=\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}$ 
Using partial fraction:
Let $\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C x+D}{\left(x^{2}+4\right)}$ .....(i)
$\Rightarrow \frac{1}{(x+1)\left(x^{2}+9\right)}$ = $\frac{(A x+B)\left(x^{2}+4\right)+(C x+D)\left(x^{2}+1\right)}{(x+1)\left(x^{2}+9\right)}$ 
$\Rightarrow$1 = (Ax + B)(x2 + 4) + (Cx + D)(x2 + 1)
$\Rightarrow$ 1 = Ax3 + 4Ax + Bx2 + 4B + Cx3 + Cx + Dx2 + D
$\Rightarrow$ 1 = (A + C)x3 +(B + D)x2 +(4A + C)x + (4B + D)
Equating the coefficients of x, x2, x3 and constant value. We get:
A + C = 0 $\Rightarrow$ C = -A
B + D = 0 $\Rightarrow$ B = -D
4A + C = 0 $\Rightarrow$ 4A = -C $\Rightarrow$ 4A = A $\Rightarrow$ 3A = 0 $\Rightarrow$ A = 0 $\Rightarrow$ C = 0
4B + D = 1 $\Rightarrow$ 4B - B = 1 $\Rightarrow$ B = $\frac{1}{3}$ $\Rightarrow$ D = $\frac{-1}{3}$
Put these values in equation (i)
$\Rightarrow \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C x+D}{\left(x^{2}+4\right)}$ 
$\Rightarrow \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=$ $\frac{(0) x+\frac{1}{3}}{\left(x^{2}+1\right)}+\frac{(0) x+\left(-\frac{1}{3}\right)}{\left(x^{2}+4\right)}$ 
$\Rightarrow \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{\frac{1}{3}}{\left(x^{2}+1\right)}+\frac{\left(-\frac{1}{3}\right)}{\left(x^{2}+4\right)}$ 
$\Rightarrow \int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x$ = $\frac{1}{3} \cdot \int \frac{1}{\left(x^{2}+1\right)} d x-\frac{1}{3} \cdot \int \frac{1}{\left(x^{2}+4\right)} d x$ 
$\Rightarrow \int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x$ = $\frac{1}{3} \cdot \int \frac{1}{\left(x^{2}+1^{2}\right)} d x-\frac{1}{3} \cdot \int \frac{1}{\left(x^{2}+2^{2}\right)} d x$ 
= $\frac{1}{3} \cdot \tan ^{-1} x-\frac{1}{3} \cdot \frac{1}{2} \tan ^{-1} \frac{x}{2}+C$ 
$\Rightarrow \mathrm{I}=\frac{1}{3} \cdot \tan ^{-1} \mathrm{x}-\frac{1}{6} \tan ^{-1} \frac{\mathrm{x}}{2}+\mathrm{C}$

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 "1 Marks" from Integrals→Integrals — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Integrate the function $\frac{x e^{x}}{(1+x)^{2}}$
→
Find:
$\int \frac{\sin^{2} \text{x} - \cos^{2} \text{x}}{\sin \text{x} \cos \text{x}} \text{dx}$
→
If $\begin{bmatrix} 9 &-1& 4\\ -2 & 1 & 3 \\ \end{bmatrix} =\text{A} + \begin{bmatrix} 1 & 2&-1 \\ 0 & 4 & 9 \\ \end{bmatrix},$then find the matrix A.
→
Classify the following as scalar and vector quantities:
Force.
→
Let X = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let R1 be a relation on X given by R1 = {(x, y ): x - y is divisible by 3} and R2 be another relation on X given by R2 = {(x, y) : {x, y) $\subset$ {1, 4, 7} or {x, y} $\subset$ {2, 5, 8} or {x, y} $\subset$ {3, 6, 9}}. Show that R1 = R2.
→
Find the direction cosines of the vector $\hat{i}+2\hat{j}+3\hat{k}.$
→
Sketch the graph of $\text{y}=\sqrt{\text{x}+1}$ in [0, 4] and determine the area of the region enclosed by the curve, the x-axis and the lines x = 0, x = 4.
→
 Write the order of the differential equation of the family of circles touching X-axis at the origin.
→
Find the length of foot of perpendicular from the point $(1,2,3)$ on plane $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=5$.
→
Show that the function given by f(x) = 7x – 3 is increasing on R.
→