Download App

Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals4 Marks
Question
Evaluate the following integrals:

$\int\frac{2\text{x}}{2+\text{x}-\text{x}^2}\text{ dx}$

✓

Answer

$\int\frac{2\text{x}\text{ dx}}{(2+\text{x}-\text{x}^2)}$

$2\text{x}=\text{A}\frac{\text{d}}{\text{dx}}\big(2+\text{x}-\text{x}^2\big)+\text{B}$

$2\text{x}=\text{A}(0+1-2\text{x})+\text{B}$

$2\text{x}=(-2\text{A})\text{x}+\text{A}+\text{B}$

Comparing the coefficients of like power of x,

$-2\text{A}=2$

$\text{A}=-1$

$\text{A}+\text{B}=0$

$-1+\text{B}=0$

$\text{B}=1$

Now, $\int\frac{2\text{x}\text{ dx}}{(2+\text{x}-\text{x}^2)}$

$=\int\Big(\frac{-1(1-2\text{x})+1}{-\text{x}^2+\text{x}+2}\Big)\text{dx}$

$=-\int\Big(\frac{1-2\text{x}}{-\text{x}^2+\text{x}+2}\Big)\text{dx}+\int\frac{\text{dx}}{-\text{x}^2+\text{x}+2}$

$=-\text{I}_1+\text{I}_2\ ....(1)$ (say) where

$\text{I}_1=\int\Big(\frac{1-2\text{x}}{-\text{x}^2+\text{x}+2}\Big)\text{dx}$

$\text{I}_2=\int\frac{\text{dx}}{-\text{x}^2+\text{x}+2}$

$\text{I}_1=\int\Big(\frac{1-2\text{x}}{-\text{x}^2+\text{x}+2}\Big)\text{dx}$

Let $-\text{x}^2+\text{x}+2=\text{t}$

$\Rightarrow(1-2\text{x})\text{dx}=\text{dt}$

$\text{I}_1=\int\frac{\text{dt}}{\text{t}}$

$\text{I}_1=\log|\text{t}|+\text{C}_1\ ....(2)$

$\text{I}_2=\int\frac{\text{dx}}{-\text{x}^2+\text{x}+2}$

$\text{I}_2=\int\frac{-\text{dx}}{\text{x}^2+\text{x}+2}$

$\text{I}_2=\int\frac{-\text{dx}}{\text{x}^2-\text{x}+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2-2}$

$\text{I}_2=\int\frac{-\text{dx}}{\big(\text{x}-\frac{1}{2}\big)^2-\big(\frac{3}{2}\big)^2}$

$\text{I}_2=-\frac{1}{2\times\frac{3}{2}}\log\Bigg|\frac{\text{x}-\frac{1}{2}-\frac{3}{2}}{\text{x}-\frac{1}{2}+\frac{3}{2}}\Bigg|+\text{C}_2$

$\text{I}_2=-\frac{1}{3}\log\Big|\frac{\text{x}-2}{\text{x}+1}\Big|+\text{C}_2\ ....(3)$

From (1) (2) and (3)

$\int\Big(\frac{2\text{x}}{2+\text{x}+\text{x}^2}\Big)\text{ dx}=-\log\big|2+\text{x}-\text{x}^2\big|-\frac{1}{3}\log\Big|\frac{\text{x}-2}{\text{x}+1}\Big|+\text{C}_1+\text{C}_2$

$=-\log\big|2+\text{x}-\text{x}^2\big|+\frac{1}{3}\log\Big|\frac{1+\text{x}}{\text{x}-2}\Big|+\text{C}$

Where, $\text{C}=\text{C}_1+\text{C}_2$

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

Similar questions

Evaluate the following definite integrals:
$\int_{0}^\limits{\infty}\frac{1}{\text{a}^2+\text{b}^2\text{x}^2} \text{ dx}$
→
If $A =\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]$, then, prove that $A^3-6 A^2+7 A+2 I=O$ and hence find $A^{-1}$.
→
Integrate the function: $\frac{x-1}{\sqrt{x^{2}-1}}$
→
Find the shortest distance between the following pairs of parallel lines whose equations are:

$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$ and $\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\big)+\mu\big(-\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$

→
Find the equation of the plane which contains two parallel lines $\frac{\text{x}-4}{1}=\frac{\text{y}-3}{-4}=\frac{\text{z}-2}{5}$ and $\frac{\text{x}-3}{1}=\frac{\text{y}+2}{-4}=\frac{\text{z}}{5}.$
→
For the matrix $\text{A}=\begin{bmatrix}1 & -1 & 1 \\2 & 3 & 0 \\ 18 & 2 & 10 \end{bmatrix},$ show that A (adjoint A) = 0.
→
Using matrices, solve the following system of equations:

3x - y + z = 5

2x - 2y + 3z = 7

x + y - z = -1.

→
Find the area enclosed by the curve y = -x2 and the strainght line x + y + 2 = 0.
→
Three dice are thrown at the sametime. Find the probability of getting three two’s, if it is known that the sum of the numbers on the dice was six.
→
Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{4}}\Big(\sqrt{\tan\text{x}}+\sqrt{\cot}\text{x}\Big)\text{dx}$
→