What is the integral from 0 to 8 of x/(x^2+1) ?

The integral from 0 to 8 of x/(x^2+1) is 1/2 ln(65)

Popular >

integral from 0 to 8 of x/(x^2+1)

\int_{0}^{8} \frac{x}{x ^{2} + 1} d x

\frac{1}{2} ln (65)

Decimal Notation

2.08719 \dots

Solution steps

\int_{0}^{8} \frac{x}{x ^{2} + 1} d x

Apply u-substitution

: \int_{1}^{65} \frac{1}{2 u} d u

= \int_{1}^{65} \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{1}^{65} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{1}^{65}

Compute the boundaries:

ln (65)

= \frac{1}{2} ln (65)

\int_{- 1}^{2} x^{3} - 3 x^{2} - x + 3 d x

\int_{0}^{t} x e^{- 2 x} d x

\int_{0}^{4} \frac{2}{( x + 2 ) ^{2}} d x

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

\int_{1}^{2} (4 - x^{2}) d x

What is the integral from 0 to 8 of x/(x^2+1) ?
The integral from 0 to 8 of x/(x^2+1) is 1/2 ln(65)