What is the integral from 0 to 1 of x/(1+4x^2) ?

The integral from 0 to 1 of x/(1+4x^2) is 1/8 ln(5)

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integral from 0 to 1 of x/(1+4x^2)

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

\frac{1}{8} ln (5)

Decimal

0.20117 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{5} \frac{1}{8 u} d u

= \int_{1}^{5} \frac{1}{8 u} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{1}^{5} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{8} [ln ∣ u ∣]_{1}^{5}

Compute the boundaries:

ln (5)

= \frac{1}{8} ln (5)

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

\int_{1}^{4} (2 - 9 x - 4 x^{2}) d x

\int_{0}^{1} x e^{- 9 x} d x

\int_{0}^{2} sin (t^{2}) d t

\int_{0}^{3} 5 (1 - e^{x}) d x

What is the integral from 0 to 1 of x/(1+4x^2) ?
The integral from 0 to 1 of x/(1+4x^2) is 1/8 ln(5)