What is the integral from 1 to 3 of x/(x^2+4) ?

The integral from 1 to 3 of x/(x^2+4) is 1/2 (ln(13)-ln(5))

Popular >

integral from 1 to 3 of x/(x^2+4)

\int_{1}^{3} \frac{x}{x ^{2} + 4} d x

\frac{1}{2} (ln (13) - ln (5))

Decimal Notation

0.47775 \dots

Solution steps

\int_{1}^{3} \frac{x}{x ^{2} + 4} d x

Apply u-substitution

: \int_{5}^{13} \frac{1}{2 u} d u

= \int_{5}^{13} \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_{5}^{13} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{5}^{13}

Compute the boundaries:

ln (13) - ln (5)

= \frac{1}{2} (ln (13) - ln (5))

\int_{4}^{4} 1

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

\int_{0}^{\frac{2}{11}} \frac{1}{121 x ^{2} + 4} d x

\int_{0}^{l n (2)} 3 e^{3 x} d x

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

What is the integral from 1 to 3 of x/(x^2+4) ?
The integral from 1 to 3 of x/(x^2+4) is 1/2 (ln(13)-ln(5))