What is the integral of (ln(1+x^2))/(x^2) ?

The integral of (ln(1+x^2))/(x^2) is -(ln(1+x^2))/x+2arctan(x)+C

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integral of (ln(1+x^2))/(x^2)

integral

\frac{ln ( 1 + x ^{2} )}{x ^{2}}

- \frac{ln ( 1 + x ^{2} )}{x} + 2 arctan (x) + C

Solution steps

\int \frac{ln ( 1 + x ^{2} )}{x ^{2}} d x

Apply Integration By Parts

: - \frac{ln ( 1 + x ^{2} )}{x} - \int - \frac{2}{1 + x ^{2}} d x

= - \frac{ln ( 1 + x ^{2} )}{x} - \int - \frac{2}{1 + x ^{2}} d x

\int - \frac{2}{1 + x ^{2}} d x = - 2 arctan (x)

= - \frac{ln ( 1 + x ^{2} )}{x} - (- 2 arctan (x))

Simplify

= - \frac{ln ( 1 + x ^{2} )}{x} + 2 arctan (x)

Add a constant to the solution

= - \frac{ln ( 1 + x ^{2} )}{x} + 2 arctan (x) + C

\frac{1}{t ^{2}}

\frac{x}{( x ^{2} + 1 ) ^{\frac{1}{3}}}

2^{2 x}

9

\frac{( sin ( x ) - 1 )}{cos ^{2} ( x )}

What is the integral of (ln(1+x^2))/(x^2) ?
The integral of (ln(1+x^2))/(x^2) is -(ln(1+x^2))/x+2arctan(x)+C