What is the integral from 0 to x of ln(1+x^2) ?

The integral from 0 to x of ln(1+x^2) is xln(x^2+1)-2(x-arctan(x))

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integral from 0 to x of ln(1+x^2)

\int_{0}^{x} ln (1 + x^{2}) d x

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

Solution steps

\int_{0}^{x} ln (1 + x^{2}) d x

Apply Integration By Parts

: [x ln (1 + x^{2}) - \int \frac{2 x ^{2}}{1 + x ^{2}} d x]_{0}^{x}

= [x ln (1 + x^{2}) - \int \frac{2 x ^{2}}{1 + x ^{2}} d x]_{0}^{x}

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

= [x ln (1 + x^{2}) - 2 (- arctan (x) + x)]_{0}^{x}

Compute the boundaries:

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

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

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What is the integral from 0 to x of ln(1+x^2) ?
The integral from 0 to x of ln(1+x^2) is xln(x^2+1)-2(x-arctan(x))