What is the integral from-3 to 3 of ln(y^2+1) ?

The integral from-3 to 3 of ln(y^2+1) is 4arctan(3)+6ln(10)-12

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integral from-3 to 3 of ln(y^2+1)

\int_{- 3}^{3} ln (y^{2} + 1) d y

4 arctan (3) + 6 ln (10) - 12

Decimal

6.81169 \dots

Solution steps

\int_{- 3}^{3} ln (y^{2} + 1) d y

Apply Integration By Parts

: [y ln (y^{2} + 1) - \int \frac{2 y ^{2}}{y ^{2} + 1} d y]_{- 3}^{3}

= [y ln (y^{2} + 1) - \int \frac{2 y ^{2}}{y ^{2} + 1} d y]_{- 3}^{3}

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

= [y ln (y^{2} + 1) - 2 (- arctan (y) + y)]_{- 3}^{3}

Compute the boundaries:

4 arctan (3) + 6 ln (10) - 12

= 4 arctan (3) + 6 ln (10) - 12

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What is the integral from-3 to 3 of ln(y^2+1) ?
The integral from-3 to 3 of ln(y^2+1) is 4arctan(3)+6ln(10)-12