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

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

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

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

2 ln (2) + π - 4

Decimal

0.52788 \dots

Solution steps

\int_{- 1}^{1} 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]_{- 1}^{1}

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

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

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

Compute the boundaries:

2 ln (2) + π - 4

= 2 ln (2) + π - 4

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\int_{0}^{x} 1 - x d x

\int_{2}^{4} \frac{m ^{2}}{m + 1} d m

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