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

The integral from 0 to 2 of ln(1+x^2) is 2ln(5)-2(-arctan(2)+2)

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

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

2 ln (5) - 2 (- arctan (2) + 2)

Decimal Notation

1.43317 \dots

Solution steps

\int_{0}^{2} 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}^{2}

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

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

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

Compute the boundaries:

2 ln (5) - 2 (- arctan (2) + 2)

= 2 ln (5) - 2 (- arctan (2) + 2)

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What is the integral from 0 to 2 of ln(1+x^2) ?
The integral from 0 to 2 of ln(1+x^2) is 2ln(5)-2(-arctan(2)+2)