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 Notation

0.26394 \dots

Solution steps

\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}

= [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)

\int \frac{1}{x ^{2} + x + 1} d x

- \frac{3 ( z ^{2} + 1 )}{z ^{2}}

y^{2} = 6 - x, 3 y = x + 12

\int_{π}^{2 π} 9 x sin (x) d x

n = 0 \sum \infty 12 \cdot 626582

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)