Question

Evaluate : $\int \frac{3x-1}{\sqrt{x^2+9}}dx$

Answer 1

$\int \frac{3x-1}{\sqrt{x^2+9}}dx$

$=\int \frac{3x}{\sqrt{x^2+9}}dx-\int \frac{1}{\sqrt{x^2+9}}dx$

$=\frac{3}{2}\int \frac{2x}{\sqrt{x^2+9}}dx-\int \frac{1}{\sqrt{x^2+9}}dx$

$=\frac{3}{2}\int \frac{2x}{\sqrt{x^2+9}}dx-\log \mid x+\sqrt{x^2+9\mid }$

$[\therefore \int \frac{1}{\sqrt{x^2+a^2}}dx=\log \mid x+\sqrt{x^2+a^2}\mid +c]$

Substituting $x^2+9=t\Rightarrow 2xdx=dt$,

$\Rightarrow \frac{3}{2}\int \frac{2x}{\sqrt{x^2+9}}dx-\log \mid x+\sqrt{x^2+9}\mid$

$=\frac{3}{2}\int \frac{1}{\sqrt{t}}dt-\log \mid x+\sqrt{x^2+9}\mid$

$=\frac{3}{2}\int t^{-\frac{1}{2}}dt-\log \mid x+\sqrt{x^2+9}\mid$

$=\frac{3}{2}\left(\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right)-\log \mid x+\sqrt{x^2+9}\mid +c$

$=\frac{3}{2}\left(\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right)-\log \mid x+\sqrt{x^2+9}\mid +c$

$=3t^{\frac{1}{2}}-\log \mid x+\sqrt{x^2+9}\mid +c$

$=3\sqrt{x^2+9}-\log \mid x+\sqrt{x^2+9}\mid +c$

$[\because x^2+9=t]$

Where $c$ is constant of integration
Hence, the required integration is
$3\sqrt{x^2+9}-\log \mid x+\sqrt{x^2+9}\mid +c$