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

The integral from 0 to 1 of x/(1+5x^2) is 1/10 ln(6)

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integral from 0 to 1 of x/(1+5x^2)

\int_{0}^{1} \frac{x}{1 + 5 x ^{2}} d x

\frac{1}{10} ln (6)

Decimal

0.17917 \dots

Solution steps

\int_{0}^{1} \frac{x}{1 + 5 x ^{2}} d x

Apply u-substitution

: \int_{1}^{6} \frac{1}{10 u} d u

= \int_{1}^{6} \frac{1}{10 u} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int_{1}^{6} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{10} [ln ∣ u ∣]_{1}^{6}

Compute the boundaries:

ln (6)

= \frac{1}{10} ln (6)

\int_{0}^{1} \frac{10}{1 + x ^{2}} d x

\int_{1}^{4} (\frac{3}{2} x^{2} - 2 x) d x

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

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

\int_{0}^{x} tan (x) d x

What is the integral from 0 to 1 of x/(1+5x^2) ?
The integral from 0 to 1 of x/(1+5x^2) is 1/10 ln(6)