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

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

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

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

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

Decimal Notation

0.23104 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{4} \frac{1}{6 u} d u

= \int_{1}^{4} \frac{1}{6 u} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int_{1}^{4} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} [ln ∣ u ∣]_{1}^{4}

Compute the boundaries:

2 ln (2)

= \frac{1}{6} \cdot 2 ln (2)

Simplify

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

\int_{5}^{\infty} e^{- x} d x

\int_{x}^{1} e^{y^{2}} d y

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

\int_{2}^{7} \frac{1}{x + 2} d x

\int_{0}^{1} p^{4} e^{- p} d p

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