What is the integral from 0 to 4 of (x^2)/(x+1) ?

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

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integral from 0 to 4 of (x^2)/(x+1)

\int_{0}^{4} \frac{x ^{2}}{x + 1} d x

ln (5) + 4

Decimal

5.60943 \dots

Solution steps

\int_{0}^{4} \frac{x ^{2}}{x + 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

Expand

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

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

\int_{1}^{5} u d u = 12

\int_{1}^{5} 2 d u = 8

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

= 12 - 8 + ln (5)

Simplify

= ln (5) + 4

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