# Get Answers to all your Questions

### Answers (1)

Answer:

$\frac{1}{\sqrt{17}}log\left ( \frac{21+5\sqrt{7}}{4} \right )$

Hint: Use indefinite formula and then put limits to solve this integral

Given:

$\int_{0}^{2}\frac{1}{4+x-x^{2}}dx$

Solution:

\begin{aligned} &\int_{0}^{2} \frac{1}{4+x-x^{2}} d x=\int_{0}^{2} \frac{1}{-\left(x^{2}-x-4\right)} d x \\ &=-\int_{0}^{2} \frac{1}{x^{2}-2 x \cdot \frac{1}{2}+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}-4} d x \\ &=-\int_{0}^{2} \frac{1}{\left(x-\frac{1}{2}\right)^{2}-\left(\frac{1}{4}\right)-4} d x \\ &=-\int_{0}^{2} \frac{1}{\left(x-\frac{1}{2}\right)^{2}-\left(\frac{1+16}{4}\right)} d x \end{aligned}

\begin{aligned} &=-\int_{0}^{2} \frac{1}{\left(x-\frac{1}{2}\right)^{2}-\left(\frac{17}{4}\right)} d x \\ &=\int_{0}^{2} \frac{1}{\left(\frac{17}{4}\right)-\left(x-\frac{1}{2}\right)^{2}} d x \end{aligned}

Putting $\left ( x-\frac{1}{2} \right )=t$

$\Rightarrow dx=dt$

When $x=0$ then

$t=-\frac{1}{2}$

and when $x=2$ then

$t=2-\frac{1}{2}=\frac{4-1}{2}=\frac{3}{2}$

Then

\begin{aligned} &\int_{0}^{2} \frac{1}{4+x-x^{2}} d x=\int_{-\frac{1}{2}}^{\frac{3}{2}} \frac{1}{\left(\frac{\sqrt{17}}{2}\right)^{2}-t^{2}} d x \\ &=\left[\frac{1}{2 \times \frac{\sqrt{17}}{2}} \log \left|\frac{\frac{\sqrt{17}}{2}+t}{\frac{\sqrt{17}}{2}-t}\right|\right]_{\frac{-1}{2}}^{\frac{3}{2}} \end{aligned}

\begin{aligned} &\left.=\frac{1}{\sqrt{17}}\left[\log \left|\frac{\frac{\sqrt{17}}{2}+\frac{3}{2}}{\frac{\sqrt{17}}{2}-\frac{3}{2}}\right|-\log \mid \frac{\frac{\sqrt{17}}{2}+\left(-\frac{1}{2}\right)}{\frac{\sqrt{17}}{2}-\left(-\frac{1}{2}\right)}\right)\right] \\ &=\frac{1}{\sqrt{17}}\left[\log \left|\frac{\frac{\sqrt{17}}{2}+\frac{3}{2}}{\frac{\sqrt{17}-3}{2}}\right|-\log \left|\frac{\frac{\sqrt{17}}{2}+\left(-\frac{1}{2}\right)}{\frac{\sqrt{17}}{2}+\frac{1}{2}}\right|\right] \end{aligned}

$=\frac{1}{\sqrt{17}}\left [ log|\frac{\sqrt{17}+3}{\sqrt{17}-3}|-log|\frac{\frac{\sqrt{17}-1}{2}}{\frac{\sqrt{17}+1}{2}}| \right ]$

$=\frac{1}{\sqrt{17}}\left[\log \left|\frac{\sqrt{17}+3}{\sqrt{17}-3}\right|-\log \left|\frac{\sqrt{17}-1}{\sqrt{17}+1}\right|\right]$

$\left[\because \log m-\log n=\log \left(\frac{m}{n}\right)\right]$

$=\frac{1}{\sqrt{17}} \log \left[\frac{\left(\frac{\sqrt{17}+3}{\sqrt{17}-3}\right)}{\left(\frac{\sqrt{17}-1}{\sqrt{17}+1}\right)}\right]$

$=\frac{1}{\sqrt{17}} \log \left(\frac{\sqrt{17}+3}{\sqrt{17}-3} \times \frac{\sqrt{17}+1}{\sqrt{17}-1}\right)$

$=\frac{1}{\sqrt{17}} \log \left(\frac{17+3 \sqrt{17}+\sqrt{17}+3}{17-3 \sqrt{17}-\sqrt{17}+3}\right)$

$=\frac{1}{\sqrt{17}} \log \left(\frac{20+4 \sqrt{17}}{20-4 \sqrt{17}}\right)$

$=\frac{1}{\sqrt{17}} \log \left(\frac{4(5+\sqrt{17})}{4(5-\sqrt{17})}\right)$

$=\frac{1}{\sqrt{17}} \log \left(\frac{5+\sqrt{17}}{5-\sqrt{17}}\right)$

$=\frac{1}{\sqrt{17}} \log \left(\frac{5+\sqrt{17}}{5-\sqrt{17}} \times \frac{5+\sqrt{17}}{5+\sqrt{17}}\right)$

$=\frac{1}{\sqrt{17}} \log \left(\frac{(5+\sqrt{17})^{2}}{(5-\sqrt{17})(5+\sqrt{17})}\right)$

$=\frac{1}{\sqrt{17}} \log \left(\frac{5^{2}+(\sqrt{17})^{2}+2.5 \cdot \sqrt{17}}{(5)^{2}-(\sqrt{17})^{2}}\right)$

$\left[\begin{array}{l}\because(a+b)^{2}=a^{2}+b^{2}+2 a b \\ \left(a+b(a-b)=\left(a^{2}-b^{2}\right)\right.\end{array}\right]$

$=\frac{1}{\sqrt{17}} \log \left(\frac{25+17+10 \sqrt{17}}{25-17}\right)$

$=\frac{1}{\sqrt{17}} \log \left(\frac{2(21+5 \sqrt{17})}{8}\right)$

$=\frac{1}{\sqrt{17}} \log \left(\frac{21+5 \sqrt{17}}{4}\right)$

View full answer

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support