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)

Popular >

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

Eliminate Absolutes

= \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_{1}^{z} y e^{y^{2} - z} d y

\int_{0}^{1} (3 x^{2} + 2 x) d x

\int_{0}^{10} sin (\frac{π x}{10}) d x

\int_{- \infty}^{0} z e^{6 z} d z

\int_{- 2}^{3} 2 x^{3} - 8 x^{2} + 2 x + 12 d x

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)