What is the integral from 0 to 0.37 of ln(1+x^2) ?

The integral from 0 to 0.37 of ln(1+x^2) is 0.01623…

Popular >

integral from 0 to 0.37 of ln(1+x^2)

\int_{0}^{0.37} ln (1 + x^{2}) d x

0.01623 \dots

Solution steps

\int_{0}^{0.37} 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}^{0.37}

= [x ln (1 + x^{2}) - \int \frac{2 x ^{2}}{1 + x ^{2}} d x]_{0}^{0.37}

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

= [x ln (1 + x^{2}) - 2 (- arctan (x) + x)]_{0}^{0.37}

Compute the boundaries:

0.01623 \dots

= 0.01623 \dots

\int_{0}^{2 π} \frac{1 + cos ( 2 x )}{2} d x

\int_{- \infty}^{- 3} \frac{x}{4 - x ^{2}} d x

\int_{0}^{4} 3 x^{3} d x

\int_{6.3}^{6.6} 12 x^{2} - 6 x + 4 d x

\int_{- 4}^{3} (- x^{2} - 3 x) d x

What is the integral from 0 to 0.37 of ln(1+x^2) ?
The integral from 0 to 0.37 of ln(1+x^2) is 0.01623…