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# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{1}{1+x-x^{2}} d x$

Solution:

: let $\mathrm{I}=\int \frac{1}{1+\mathrm{x}-\mathrm{x}^{2}} \mathrm{~d} \mathrm{x}=\int \frac{1}{-\left(\mathrm{x}^{2}-\mathrm{x}-1\right)} \mathrm{dx}$

$=\int \frac{1}{-\left(x^{2}-x-1\right)} d x$

$=\int \frac{1}{-\left(x^{2}-x-\frac{1}{4}-1+\frac{1}{4}\right)} d x$

$=\int \frac{1}{-\left(\left(x-\frac{1}{2}\right)^{2}-\frac{5}{4}\right)} d x$

$=\int \frac{1}{\left(\left(\frac{\sqrt{5}}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}\right)} d x$

$I=\frac{1}{2 \times \frac{\sqrt{5}}{2}} \log \left|\frac{\frac{\sqrt{5}}{2}+\left(x-\frac{1}{2}\right)}{\frac{\sqrt{5}}{2}-\left(x-\frac{1}{2}\right)}\right|+c$

[since, $\left.\int \frac{1}{x^{2}-(a)^{2}} d x=\frac{1}{2 \times a} \log \left|\frac{x-a}{x+a}\right|+c\right]$

$I=\frac{1}{\sqrt{5}} \log \left|\frac{\sqrt{5}+2 x-1}{\sqrt{5}-2 x+1}\right|+c$

$I=\frac{1}{\sqrt{5}} \log \left|\frac{\sqrt{5}-1+2 x}{\sqrt{5}+1-2 x}\right|+c$