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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

\int_{1}^{2} 1 d u = 1

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

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

\int \frac{1}{5} x d x

\int \frac{2}{x ^{3} x ^{4} - 81} d x

\frac{d}{d t} (e^{- a t} cos (wt))

\int \frac{x ^{2} 2 ^{x} + x}{x ^{2}} d x

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

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