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

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

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

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

ln (3)

Decimal Notation

1.09861 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{3} \frac{( u - 1 ) ^{2}}{u} d u

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

Expand

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

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

\int_{1}^{3} u d u = 4

\int_{1}^{3} 2 d u = 4

\int_{1}^{3} \frac{1}{u} d u = ln (3)

= 4 - 4 + ln (3)

Simplify

= ln (3)

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

y = \frac{x ^{3}}{6} + \frac{1}{2 x}

x - cos (x)

\int_{0}^{2} 6 t d t

\frac{\partial}{\partial x} (\frac{1}{x - 3})

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