Evaluate the following intregals: x x^2+6x+10 dx - Vidyadip

Question

Evaluate the following intregals:
$\int\frac{\text{x}}{\sqrt{\text{x}^2+6\text{x}+10}}\ \text{dx}$

✓

Answer

Let $\text{I}\int\frac{\text{x}}{\sqrt{\text{x}^2+6\text{x}+10}}\ \text{dx}$
Let $\text{x}=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+6\text{x}+10)+\mu$
$=\lambda(2\text{x}+6)+\mu$
$\text{x}=(2\lambda)\text{x}+6\lambda+\mu$
Comparing the co-efficient of like powers of x.
$2\lambda=1\Rightarrow\lambda=\frac{1}{2}$
$6\lambda+\mu=0\Rightarrow6\Big(\frac{1}{2}\Big)+\mu=0$
$\mu=-3$
So, $\text{I}_1=\int\frac{\frac{1}{2}(2\text{x}+6)=3}{\sqrt{\text{x}^2+6\text{x}+10}}\ \text{dx}$
$\frac{1}{2}\int\frac{{2\text{x}+6}=3}{\sqrt{\text{x}^2+6\text{x}+10}}\ \text{dx}-3\text{ I }\frac{1}{\sqrt{\text{x}^2+2\text{x}(3)+(3)^2+10}}$
$\text{I}_1=\frac{1}{2}\int\frac{2\text{x}+6}{\sqrt{\text{x}^2+6\text{x}+10}}\ \text{dx}-3\int\frac{1}{\sqrt{(\text{x}+3)^2+(1)^2}}\text{dx}$
$\text{I}=\sqrt{\text{x}^2+6\text{x}+10}-3\log\big|\text{x}+3+\sqrt{\text{x}^2+6\text{x}+10}\big|+\text{c}$
$\text{I}_1=\frac{1}{2}(2\sqrt{\text{x}^2+6\text{x}+10})-3\log\big|\text{x}+3+\sqrt{(\text{x}+3)^2+1}\big|+\text{c}$ $\Big[\text{since},\int\frac{1}{\sqrt{\text{x}}}\text{dx}-2\sqrt{\text{x}}+\text{c},\int\frac{1}{\sqrt{\text{x}^2+\text{a}^2}}\text{dx}-\log\big|\text{x}+\sqrt{\text{x}}^2+\text{a}^2\big|+\text{c}\Big]$
$\text{I}=\sqrt{\text{x}^2+6\text{x}+10}-3\log\big|\text{x}+3+\sqrt{\text{x}^2+6\text{x}+10}\big|+\text{c}$

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

Evaluate the following integrals:
$\int_{0}^\limits{\pi}\frac{\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$

A rectangular sheet of tin $45 \ cm$ by $24 \ cm$ is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?

Find the intervals in which the following functions are increasing or decreasing. $f(x) = x^{2 }+ 2x - 5$

Show that the following system of linear equations is consistent and also find solution:
$2x + 2y − 2z = 1$
$4x + 4y − z = 2$
$6x + 6y + 2z = 3$

If $\text{f(x)}=\sqrt{1-\text{x}}$ and $\text{g(x)}=\log_\text{e}\text{x}$ are two real functions, then describe, functions fog and gof.

The slope of the tangent at a point P(x, y) on a curve is $\frac{-\text{x}}{\text{y}}$. If the curve passs es through the point (3, -4). Find the equation of the curve.

Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}}+\cos\text{x},&\text{if }\text{ x}\neq0\\5,&\text{if }\text{ x}=0\end{cases}$

A factory manufactures two types of screws A and B, each type requiring the use of two machines, an automatic and a hand-operated. It takes 4 minutes on the automatic and 6 minutes on the hand-operated machines to manufacture a packet of screws ‘A’ while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture a packet of screws ‘B’. Each machine is available for at most 4 hours on any day. The manufacturer can sell a packet of screws ‘A’ at a profit of 70 paise and screws ‘B’ at a profit of Rs. 1. Assuming that he can sell all the screws he manufactures, how many packets of each type should the factory owner produce in a day in order to maximize his profit? Formulate the above LPP and solve it graphically and find the maximum profit.

Verify Rolle's theorem for the following function on the indicated intervals $f(x) = x^2+ 5 x + 6$ on the interval $[-3, -2]$

Find the shortest distance between the following pairs of lines whose cartesian equation are:
$\frac{\text{x}-1}{-1}=\frac{\text{y}+2}{1}=\frac{\text{z}-3}{-2}$ and $\frac{\text{x}-1}{1}=\frac{\text{y}+1}{2}=\frac{\text{z}+1}{-2}$