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

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

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

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

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

Decimal Notation

0.19314 \dots

Solution steps

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

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}

x y^{2} d x = d y - x d x, y (0) = - 1

\frac{d x}{d y}

\int \frac{x ^{2}}{1 - x} d x

θ^{^{''}} + \frac{4 k}{3} θ = 0

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

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