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

The integral from 0 to 1 of ln(1+x^2) is ln(2)-2(-pi/4+1)

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

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

ln (2) - 2 (- \frac{π}{4} + 1)

Decimal Notation

0.26394 \dots

Solution steps

\int_{0}^{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]_{0}^{1}

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

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

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

Compute the boundaries:

ln (2) - 2 (- \frac{π}{4} + 1)

= ln (2) - 2 (- \frac{π}{4} + 1)

1 + x, 1

\int \frac{1}{4} cos^{2} (2 x) d x

y^{^{'}} + 0.2 \cdot y = 5

f (x) = x^{\frac{1}{6}}

\int x (x^{2} + 2) d x

What is the integral from 0 to 1 of ln(1+x^2) ?
The integral from 0 to 1 of ln(1+x^2) is ln(2)-2(-pi/4+1)