Question

Find
$\int \frac{dt}{\sqrt{t}}+\left(3-\frac{4}{3}\right)\int \frac{dx}{\sqrt{3x^2+4x+1}}$

Answer 1

Consider the given integral.

$I=\int \frac{dt}{\sqrt{t}}+(3-\frac{4}{3})\int \frac{dx}{\sqrt{3x^2+4x+1}}$

$I=\int \frac{dt}{\sqrt{t}}+\frac{5}{3}\int \frac{dx}{\sqrt{3x^2+4x+1}}$

$I=\int \frac{dt}{\sqrt{t}}+\frac{5}{3\sqrt{3}}\int \frac{dx}{\sqrt{x^2+\frac{4}{\sqrt{3}}x+\frac{1}{\sqrt{3}}}}$

$I=\int \frac{dt}{\sqrt{t}}+\frac{5}{3\sqrt{3}}\int \frac{dx}{\sqrt{x^2+\frac{4}{\sqrt{3}}x+{\left(\frac{2}{\sqrt{3}}\right)}^2-{\left(\frac{2}{\sqrt{3}}\right)}^2+\frac{1}{\sqrt{3}}}}$

$I=\int \frac{dt}{\sqrt{t}}+\frac{5}{3\sqrt{3}}\int \frac{dx}{\sqrt{{(x+\frac{2}{\sqrt{3}})}^2-\frac{4}{3}+\frac{1}{\sqrt{3}}}}$

$I=\int \frac{dt}{\sqrt{t}}+\frac{5}{3\sqrt{3}}\int \frac{dx}{\sqrt{{(x+\frac{2}{\sqrt{3}})}^2+\frac{1}{\sqrt{3}}-\frac{4}{3}}}$

$I=\int \frac{dt}{\sqrt{t}}+\frac{5}{3\sqrt{3}}\int \frac{dx}{\sqrt{{(x+\frac{2}{\sqrt{3}})}^2+{\left(\sqrt{\frac{3-4\sqrt{3}}{3\sqrt{3}}}\right)}^2}}$

We know that

$\int \frac{dx}{\sqrt{x^2+a^2}}=\log |x+\sqrt{x^2+a^2}|+C$

Therefore,

$I=\int \frac{dt}{\sqrt{t}}+\frac{5}{3\sqrt{3}}\int \frac{dx}{\sqrt{{(x+\frac{2}{\sqrt{3}})}^2+{\left(\sqrt{\frac{3-4\sqrt{3}}{3\sqrt{3}}}\right)}^2}}$

$I=2\sqrt{t}+\frac{5}{3\sqrt{3}}\left[\log |x+\sqrt{{(x+\frac{2}{\sqrt{3}})}^2+{\left(\sqrt{\frac{3-4\sqrt{3}}{3\sqrt{3}}}\right)}^2}|\right]+C$

$I=2\sqrt{t}+\frac{5}{3\sqrt{3}}\left[\log |x+\sqrt{x^2+\frac{4}{3}x+\frac{1}{3}}|\right]+C$

$I=2\sqrt{t}+\frac{5}{3\sqrt{3}}\left[\log |x+\frac{1}{\sqrt{3}}\sqrt{3x^2+4x+1}|\right]+C$

Hence, this is the answer.