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

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

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

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

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

Solution steps

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

Expand

\frac{( u - 1 ) ^{2}}{u} : u - 2 + \frac{1}{u}

= \int_{1}^{2} u - 2 + \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

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

\int_{1}^{2} u d u = \frac{3}{2}

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

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

= \frac{3}{2} - 2 + ln (2)

Combine the fractions

- 2 + \frac{3}{2} : - \frac{1}{2}

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

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

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\int_{- 7}^{5} ∣ x ∣ d x

\int_{0}^{4} x^{2} + 9 d x

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