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

The integral from 0 to 4 of (x^2)/(x+4) is 16ln(2)-8

Popular >

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

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

16 ln (2) - 8

Decimal Notation

3.09035 \dots

Solution steps

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

Apply u-substitution

: \int_{4}^{8} \frac{( u - 4 ) ^{2}}{u} d u

= \int_{4}^{8} \frac{( u - 4 ) ^{2}}{u} d u

Expand

\frac{( u - 4 ) ^{2}}{u} : u - 8 + \frac{16}{u}

= \int_{4}^{8} u - 8 + \frac{16}{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_{4}^{8} u d u - \int_{4}^{8} 8 d u + \int_{4}^{8} \frac{16}{u} d u

\int_{4}^{8} u d u = 24

\int_{4}^{8} 8 d u = 32

\int_{4}^{8} \frac{16}{u} d u = 16 ln (2)

= 24 - 32 + 16 ln (2)

Simplify

= 16 ln (2) - 8

\int_{1}^{2} 4 x^{3} e^{x^{4}} d x

\int_{0}^{r} π (r^{2} - x^{2}) d x

\int_{0}^{2 π} 1 cos^{2} (x) d x

\int_{- a}^{a} 4 d x

\int_{1}^{3} 4 x - 3 - x^{2} d x

What is the integral from 0 to 4 of (x^2)/(x+4) ?
The integral from 0 to 4 of (x^2)/(x+4) is 16ln(2)-8