What is the integral from 0 to 0.41 of ln(1+x^2) ?

The integral from 0 to 0.41 of ln(1+x^2) is 0.02189…

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integral from 0 to 0.41 of ln(1+x^2)

\int_{0}^{0.41} ln (1 + x^{2}) d x

0.02189 \dots

Solution steps

\int_{0}^{0.41} 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}^{0.41}

= [x ln (1 + x^{2}) - \int \frac{2 x ^{2}}{1 + x ^{2}} d x]_{0}^{0.41}

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

= [x ln (1 + x^{2}) - 2 (- arctan (x) + x)]_{0}^{0.41}

Compute the boundaries:

0.02189 \dots

= 0.02189 \dots

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What is the integral from 0 to 0.41 of ln(1+x^2) ?
The integral from 0 to 0.41 of ln(1+x^2) is 0.02189…