Evaluate the following intregals: x^2+1 (2x+1)(x^2-1) dx - Vidyadip

Question

Evaluate the following intregals:
$\int\frac{\text{x}^2+1}{(2\text{x}+1)(\text{x}^2-1)}\text{ dx}$

✓

Answer

We have,$\text{I}=\int\frac{\text{x}^2+1}{(2\text{x}+1)(\text{x}^2-1)}$
$\text{I}=\int\frac{(\text{x}^2+1)\text{dx}}{(2\text{x}+1)(\text{x}^2-1)(\text{x}+1)}$
Let $\text{I}=\int\frac{(\text{x}^2+1)}{(2\text{x}+1)(\text{x}^2-1)(\text{x}+1)}=\frac{\text{A}}{2\text{x}+1}+\frac{\text{B}}{\text{x}-1}+\frac{\text{C}}{\text{x}+1}$
$\Rightarrow\int\frac{(\text{x}^2+1)}{(2\text{x}+1)(\text{x}^2-1)(\text{x}+1)}=\frac{\text{A}(\text{x}^2-1)+\text{B}(2\text{x}+1)(\text{x}+1)+\text{C}(2\text{x}+1)(\text{x}-1)}{(2\text{x}-1)(\text{x}-1)(\text{x}+1)}$
$\Rightarrow\text{x}^2+1=\text{A}(\text{x}^2-1)+\text{B}(2\text{x}+1)(\text{x}+1)\\+\text{C}(2\text{x}+1)(\text{x}-1)$
Putting x - 1 = 0
⇒ x = 1
1 + 1 = A × 0 + B × 0 + C (-2 + 1) (-1 - 1)
⇒ 2 = B (3) (2)
$\Rightarrow\text{B}=\frac{1}{3}$
Putting x + 1 = 0
⇒ x = -1
1 + 1 = A × 0 + B (-2 + 1)(-1 - 1)
⇒ 2 = C (-1) (-2)
⇒ C = 1
Putting 2x + 1 = 0
$\Rightarrow\text{x}=-\frac{1}{2}$
$\Big(-\frac{1}{2}\Big)^2+1=\text{A}\Big(\frac{1}{4}-1\Big)$
$\Rightarrow\frac{1}{4}+1=\text{A}\Big(-\frac{3}{4}\Big)$
$\Rightarrow\frac{5}{4}=\text{A}\Big(-\frac{3}{4}\Big)$
$\text{A}=-\frac{5}{3}$
$\therefore\text{I}=-\frac{5}{3}\int\frac{\text{dx}}{2\text{x}+1}+\frac{1}{3}\int\frac{\text{dx}}{\text{x}-1}+\int\frac{\text{dx}}{\text{x}+1}$
$=-\frac{5}{3}\times\frac{\log|2\text{x}+1|}{2}+\frac{1}{3}\log|\text{x}-1|+\log|\text{x}+1|+\text{C}$
$=-\frac{5}{6}\log|2\text{x}-1|+\frac{1}{3}\log|\text{x}-1|+\log|\text{x}+1|+\text{C}$

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 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

Evalute the following integrals:
$\int\frac{1-\cot\text{x}}{1+\cot\text{x}}\text{dx}$

If $\text{y}=1+\frac{\alpha}{\big(\frac{1}{\text{x}}-\alpha\big)}+\frac{\frac{\beta}{\text{x}}}{\big(\frac{1}{\text{x}}-\alpha\big)\big(\frac{1}{\text{x}}-\beta\big)}+\frac{\frac{\gamma}{\text{x}^2}}{\big(\frac{1}{\text{x}}-\alpha\big)\big(\frac{1}{\text{x}}-\beta\big)\big(\frac{1}{\text{x}}-\gamma\big)},$ find $\frac{\text{dy}}{\text{dx}}$

Solve the following differential equations:
$2(\text{y}+3)-\text{xy}\frac{\text{dy}}{\text{dx}}=0,\text{y}(1)=-2$

If $\text{x}=2\cos\text{t}-\cos2\text{t},\text{y}=2\sin\text{t}-\sin2\text{t},$ find $\frac{\text{d}^2\text{y}}{\text{dx}^2}\ \text{at}\ \text{t}=\frac{\pi}{2}.$

find the area of the region common to the circle $x^2 + y^2 = 16$ and the parabola $y^2 = 6x$.

A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs. 6,000. Three times the award money for Hard work added to that given for honesty amounts to Rs. 11,000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.

Sketch the graph y = |x - 5|. Evaluate $\int\limits_{0}^{1}|\text{x}-5|\text{dx} $ . What does this value of the integral represent on the graph.

Find the vector equation of the plane passing through three points with position vectors $\hat{\text{i}} + \hat{\text{j}} - 2\hat{\text{k}} , 2\hat{\text{i}} - \hat{\text{j}} + \hat{\text{k}}\text{ and } \hat{\text{i}} + 2\hat{\text{j}} + \hat{\text{k}}.$Also find the coordinates of the point of intersection of this plane and the line $\overrightarrow{\text{r}} = 3\hat{\text{i}} - \hat{\text{j}} - \hat{\text{k}} + \lambda(2\hat{\text{i}}- 2 \hat{\text{j}} + \hat{\text{k}}).$

Solve the following determinant equations:
$\begin{vmatrix}3&-2&\sin(3\theta)\\-7&8&\cos(2\theta)\\-11&14&2\end{vmatrix}=0$

The probability distribution of a random variable x is given as under:
$\text{P}(\text{X}=\text{x})=\begin{cases}\text{k}\text{x}^2 & \text{for}\text{ x}= 1,2,3\\2\text{kx} & \text{for}\text{ x } =4,5,6\\0&\text{otherwise} \end{cases}$
where k is a constant. Calculate

$\text{E}(\text{X})$
$\text{E}(3\text{X}^2)$
$\text{P}(\text{X}\geq4)$