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  • Integrate the functions in Exercises 1 to 24.

Integrate the functions in Exercises 1 to 24.

    Q24.    \frac{\sqrt{x^2 + 1}[\log(x^2+1)-2\log x]}{x^4}

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\frac{\sqrt{x^2 + 1}[\log(x^2+1)-2\log x]}{x^4}

Here let's first reduce the log function.

=\frac{\sqrt{x^2+1}}{x^4}\left [ \log (x^2+1)-\log x^2 \right ]dx

=\frac{\sqrt{x^2\left ( 1+\frac{1}{x^2} \right )}}{x^4}\left [ \log\frac{ (x^2+1)}{x^2} \right ]dx

=\int\frac{\sqrt{\left ( 1+\frac{1}{x^2} \right )}}{x^3}\left [ \log\left ( 1+\frac{1}{x^2} \right ) \right ]dx

Now, let 

t=1+\frac{1}{x^2}

dt=\frac{-2}{x^3}dx

So our function in terms if new variable t is :

I=\frac{-1}{2}\int \left [\log t \right ]\cdot t^{\frac{1}{2}}dt

now let's solve this By using integration by parts

I=\frac{-1}{2}\int \left [(\log t)\frac{t^\frac{3}{2}}{\frac{3}{2}} -\int \frac{1}{t}\cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}dt\right ]

I=\frac{-1}{3}t^\frac{3}{2}\log t+\frac{1}{3}\int t^{\frac{1}{2}}dt

I=\frac{-1}{3}t^\frac{3}{2}\log t+\frac{1}{3}\frac{t^{\frac{3}{2}}}{\frac{3}{2}}

I=\frac{2}{9}t^{\frac{3}{2}}-\frac{1}{3}t^{\frac{3}{2}}logt+c

I=\frac{1}{3}t^{\frac{3}{2}}\left [ \frac{2}{3}-\log t \right ]+c

I=\frac{1}{3}\left ( 1+\frac{1}{x^2} \right )^{\frac{3}{2}}\left [ \frac{2}{3}-\log \left ( 1+\frac{1}{x^2} \right ) \right ]+c

 

Posted by

Pankaj Sanodiya

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