Question
Evaluate the following integrals:
$\int\frac{\text{x}}{(\text{x}^2+1)\sqrt{\text{x}}}\text{ dx}$
$\int\frac{\text{x}}{(\text{x}^2+1)\sqrt{\text{x}}}\text{ dx}$
✓
Answer
Let $\text{I}=\int\frac{\text{x}}{(\text{x}^2+1)\sqrt{\text{x}}}\text{ dx}$
Let $\text{x}=\text{t}^2$
$\text{dx}=2\text{t dt}$
$\therefore\ 2\int\frac{\text{t dt}}{(\text{t}^2+1)\text{t}}$
$=2\Big|\frac{\text{dt}}{\text{t}^4+1}\Big|$
Dividing numerator and denominator by t2
$\text{I}=2\int\frac{\frac{\text{t}}{\text{t}^2}}{\big(\text{t}^2+\frac{1}{\text{t}^2}\big)}\text{ dt}$
$=\int\frac{\Big(1+\frac{1}{\text{t}^2}\big)-\Big(1-\frac{1}{\text{t}^2}\Big)}{\Big(\text{t}^2+\frac{1}{\text{t}^2}\Big)}\text{ dt}$
Let $\text{t}-\frac{1}{\text{t}}=\text{z}$
$\Big(1+\frac{1}{\text{t}^2}\Big)\text{ dt}=\text{dz}$ [For Ist part]
and, $\text{t}+\frac{1}{\text{t}}=\text{y}$
$\Big(1+\frac{1}{\text{t}^2}\Big)\text{ dt}=\text{dy}$ [For IInd part]
$\therefore\ \text{I}=\int\frac{\text{dz}}{\text{z}^2+2}-\int\frac{\text{dy}}{\text{y}^2-2}$
$=\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\text{z}}{\sqrt{2}}\Big)-\frac{1}{2\sqrt{2}}\log\bigg|\frac{\text{y}-\sqrt{2}}{\text{y}+\sqrt{2}}\bigg|+\text{C}$
$=\frac{1}{\sqrt{2}}\tan^{1}\Big(\frac{\text{t}^2-1}{\sqrt{2}\text{t}}\Big)-\frac{1}{2\sqrt{2}}\log\bigg|\frac{\text{x}+1-\sqrt{2\text{x}}}{\text{x}+1+\sqrt{2\text{x}}}\bigg|+\text{C}$
Let $\text{x}=\text{t}^2$
$\text{dx}=2\text{t dt}$
$\therefore\ 2\int\frac{\text{t dt}}{(\text{t}^2+1)\text{t}}$
$=2\Big|\frac{\text{dt}}{\text{t}^4+1}\Big|$
Dividing numerator and denominator by t2
$\text{I}=2\int\frac{\frac{\text{t}}{\text{t}^2}}{\big(\text{t}^2+\frac{1}{\text{t}^2}\big)}\text{ dt}$
$=\int\frac{\Big(1+\frac{1}{\text{t}^2}\big)-\Big(1-\frac{1}{\text{t}^2}\Big)}{\Big(\text{t}^2+\frac{1}{\text{t}^2}\Big)}\text{ dt}$
Let $\text{t}-\frac{1}{\text{t}}=\text{z}$
$\Big(1+\frac{1}{\text{t}^2}\Big)\text{ dt}=\text{dz}$ [For Ist part]
and, $\text{t}+\frac{1}{\text{t}}=\text{y}$
$\Big(1+\frac{1}{\text{t}^2}\Big)\text{ dt}=\text{dy}$ [For IInd part]
$\therefore\ \text{I}=\int\frac{\text{dz}}{\text{z}^2+2}-\int\frac{\text{dy}}{\text{y}^2-2}$
$=\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\text{z}}{\sqrt{2}}\Big)-\frac{1}{2\sqrt{2}}\log\bigg|\frac{\text{y}-\sqrt{2}}{\text{y}+\sqrt{2}}\bigg|+\text{C}$
$=\frac{1}{\sqrt{2}}\tan^{1}\Big(\frac{\text{t}^2-1}{\sqrt{2}\text{t}}\Big)-\frac{1}{2\sqrt{2}}\log\bigg|\frac{\text{x}+1-\sqrt{2\text{x}}}{\text{x}+1+\sqrt{2\text{x}}}\bigg|+\text{C}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Find the position vector of the food of perpendicular and the perpendicular distance from the point P with position vector $2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$ to the plane $\vec{\text{r}}.(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}})-26=0.$ Also find image or P in the plane.
→Solve the system of linear equation, using matrix method x - y + z = 4; 2x + y - 3z = 0; x + y + z = 2
A and B throw a pair of dice alternately. A wins the game if he gets a total of 6 and B wins if she gets a total of 7. It A starts the game, find the probability of winning the game by A in third throw of the pair of dice.
Maximum Z = 2x + 4y Subject to
→$\text{x}+\text{y}\geq8$
$\text{x}+4\text{y}\geq12$
$\text{x}\geq3,\text{y}\geq2$
Find $\frac{\text{dy}}{\text{dx}}$
y = ex + 10x + xx
→y = ex + 10x + xx
In Figure ABCD is a regular hexagon, which vectors are:
- Collinear.
- Equal.
- Co-initial.
- Collinear but not equal.
Evaluate the following integrals:
→$\int\frac{\text{x}^2}{\text{x}^2+7\text{x}+10}\text{ dx}$
Discuss the continuity of the function f, where f is defined by
$\text{f(x)}= \begin{cases}\ 3,\ \ \text{if}\ 0\leq \text{x}\leq 1 \\4,\ \ \text{if}\ 1<\text{x}<3\\5,\ \text{if}\ 3\leq\text{x}\leq10\end{cases}$
→$\text{f(x)}= \begin{cases}\ 3,\ \ \text{if}\ 0\leq \text{x}\leq 1 \\4,\ \ \text{if}\ 1<\text{x}<3\\5,\ \text{if}\ 3\leq\text{x}\leq10\end{cases}$
Find the equation of the curve which passes through the point $(1, \frac{\pi}{4})$ and tangent at any point 0f which makes an angle $\tan^{-1}\Big(\frac{\text{y}}{\text{x}}-\cos^{2}\frac{\text{y}}{\text{x}}\Big)$ with x-axis.
→Solve the following differential equations:
→$\frac{\text{dy}}{\text{dx}}=\text{y}\tan\text{ x, y}(0)=1$