What is the integral from 0 to 5 of x/(x^2+2) ?

The integral from 0 to 5 of x/(x^2+2) is 1/2 (3ln(3)-ln(2))

Popular >

integral from 0 to 5 of x/(x^2+2)

\int_{0}^{5} \frac{x}{x ^{2} + 2} d x

\frac{1}{2} (3 ln (3) - ln (2))

Decimal

1.30134 \dots

Solution steps

\int_{0}^{5} \frac{x}{x ^{2} + 2} d x

Apply u-substitution

: \int_{2}^{27} \frac{1}{2 u} d u

= \int_{2}^{27} \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_{2}^{27} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{2}^{27}

Compute the boundaries:

3 ln (3) - ln (2)

= \frac{1}{2} (3 ln (3) - ln (2))

\int_{0}^{2} y - \frac{y ^{2}}{4} d y

\int_{1}^{2} \frac{x ^{4} + 1}{x ^{5} + 5 x} d x

\int_{- 2}^{2} 4 - x^{2} + x^{3} d x

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

\int_{- 1}^{x} t^{5} d t

What is the integral from 0 to 5 of x/(x^2+2) ?
The integral from 0 to 5 of x/(x^2+2) is 1/2 (3ln(3)-ln(2))