What is the integral from 0 to 3 of x/(1+8x^2) ?

The integral from 0 to 3 of x/(1+8x^2) is 1/16 ln(73)

Popular >

integral from 0 to 3 of x/(1+8x^2)

\int_{0}^{3} \frac{x}{1 + 8 x ^{2}} d x

\frac{1}{16} ln (73)

Decimal

0.26815 \dots

Solution steps

\int_{0}^{3} \frac{x}{1 + 8 x ^{2}} d x

Apply u-substitution

: \int_{1}^{73} \frac{1}{16 u} d u

= \int_{1}^{73} \frac{1}{16 u} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int_{1}^{73} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{16} [ln ∣ u ∣]_{1}^{73}

Compute the boundaries:

ln (73)

= \frac{1}{16} ln (73)

\int_{- π}^{π} 0 d x

\int_{0}^{\frac{π}{2}} e^{x} d x

\int_{0}^{π} t^{2} sin (n t) d t

\int_{2}^{4} 5^{4 x} d x

\int_{1}^{4} 8 x^{3} - 2 x + e^{x} d x

What is the integral from 0 to 3 of x/(1+8x^2) ?
The integral from 0 to 3 of x/(1+8x^2) is 1/16 ln(73)