Question

Evaluate the following integral:

$\int \frac{2x-4}{\sqrt{x^2-4x-5}}dx+\int \frac{6}{\sqrt{(x-2{)}^2-9}}=$

• A. $\sqrt{x^2-4x-5}+\log |x+\sqrt{x^2-4x-5}|+c$
• B. $\log \left|\sqrt{x^2-4x-5}\right|+\sqrt{x^2-4x-5}+c$
• C. $\sqrt{x^2-4x-5}+6\log \left|\right(x-2)x+\sqrt{x^2-4x-5}|+c$
• D. $2\sqrt{x^2-4x-5}+6\log \left|\right(x-2)+\sqrt{x^2-4x-5}|+c$

Answer 1

The correct option is D $2\sqrt{x^2-4x-5}+6\log \left|\right(x-2)+\sqrt{x^2-4x-5}|+c$

Consider, $I=\int \frac{(2x-4)}{\sqrt{x^2-4x-5}}dx+\int \frac{6}{\sqrt{{(x-2)}^2-9}}dx$

Let, $I=I_1+I_2$

Now,

$I_1=\int \frac{(2x-4)}{\sqrt{{(x-2)}^2-9}}dx-\left(1\right)$

Let $z^2={(x-2)}^2-9$

Differentiating wrt $x$

$2zdz=2(x-2)dx$

Substituting in $\left(1\right)$

$I_1=\int \frac{2zdz}{z}=2\int 1.dz=2z+c$

$\Rightarrow$ $I_1=2\sqrt{x^2-4x-5}+c$

Now,

$I_2=\int \frac{6}{\sqrt{{(x-2)}^2-9}}dx$

Let $z=\sqrt{{(x-2)}^2-9}+(x-2)$

Let us consider a function $y=x+\sqrt{x^2-a^2}$

Differentiating wrt $x$

$y^{\prime }=1+\frac{x}{\sqrt{x^2-a^2}}=\frac{x+\sqrt{x^2-a^2}}{\sqrt{x^2-a^2}}$

Similarly differentiating wrt $z$

$dz=\frac{(x-2)+\sqrt{{(x-2)}^2-9}}{\sqrt{{(x-2)}^2-9}}dx$

$\Rightarrow$ $I_2=6\int \left(\frac{1}{\sqrt{{(x-2)}^2-9}}\times \frac{(x-2)+\sqrt{{(x-2)}^2-9}}{(x-2)+\sqrt{{(x-2)}^2-9}}\right)dx$

$I_2=6\int \frac{dz}{z}=6\log z+c$

$\Rightarrow$ $I_2=6\log |(x-2)+\sqrt{x^2-4x-5}|+c$

$\Rightarrow$ $I=I_1+I_2$

$\Rightarrow$ $I=2\sqrt{x^2-4x-5}+6\log |(x-2)+\sqrt{x^2-4x-5}|+c$