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

The integral from 0 to 1 of ln|x^2+1| is ln(2)-2(-pi/4+1)

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integral from 0 to 1 of ln|x^2+1|

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

ln (2) - 2 (- \frac{π}{4} + 1)

Decimal

0.26394 \dots

Solution steps

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

Eliminate Absolutes

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

Apply Integration By Parts

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

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

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

Compute the boundaries:

ln (2) - 2 (- \frac{π}{4} + 1)

= ln (2) - 2 (- \frac{π}{4} + 1)

What is the integral from 0 to 1 of ln|x^2+1| ?
The integral from 0 to 1 of ln|x^2+1| is ln(2)-2(-pi/4+1)