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

The integral from 0 to 2 of (x^3)/(x^2+1) is 1/2 (4-ln(5))

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

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

\frac{1}{2} (4 - ln (5))

Decimal Notation

1.19528 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{5} \frac{u - 1}{2 u} d u

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

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= \frac{1}{2} \cdot \int_{1}^{5} 1 - \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{1}^{5} 1 d u - \int_{1}^{5} \frac{1}{u} d u)

\int_{1}^{5} 1 d u = 4

\int_{1}^{5} \frac{1}{u} d u = ln (5)

= \frac{1}{2} (4 - ln (5))

\frac{\partial}{\partial x} (\frac{yz}{x ^{2} + y ^{2}})

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

\frac{x ^{6} - 1}{x - 1}

\int \frac{1}{2 x ^{2} + 4 x + 5} d x

\frac{d}{d x} (\frac{x ^{2} + 6}{sin ^{4} ( x )})

What is the integral from 0 to 2 of (x^3)/(x^2+1) ?
The integral from 0 to 2 of (x^3)/(x^2+1) is 1/2 (4-ln(5))