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

The integral from 0 to 3 of x/(1+x^2) is 1/2 ln(10)

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

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

\frac{1}{2} ln (10)

Solution steps

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

Apply u-substitution

: \int_{1}^{10} \frac{1}{2 u} d u

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{1}^{10}

Compute the boundaries:

ln (10)

= \frac{1}{2} ln (10)

\int_{2}^{7} 700 x 3 + 2 x d x

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

\int_{3}^{1} 1

\int_{0}^{4} (11 - x) x d x

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

What is the integral from 0 to 3 of x/(1+x^2) ?
The integral from 0 to 3 of x/(1+x^2) is 1/2 ln(10)