What is the integral of (x^2)/(x+8) ?

The integral of (x^2)/(x+8) is ((x+8)^2)/2-16(x+8)+64ln|x+8|+C

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integral of (x^2)/(x+8)

\int \frac{x ^{2}}{x + 8} d x

\frac{( x + 8 ) ^{2}}{2} - 16 (x + 8) + 64 ln ∣ x + 8 ∣ + C

Solution steps

\int \frac{x ^{2}}{x + 8} d x

Apply u-substitution

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

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

Expand

\frac{( u - 8 ) ^{2}}{u} : u - 16 + \frac{64}{u}

= \int u - 16 + \frac{64}{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 u d u - \int 16 d u + \int \frac{64}{u} d u

\int u d u = \frac{u ^{2}}{2}

\int 16 d u = 16 u

\int \frac{64}{u} d u = 64 ln ∣ u ∣

= \frac{u ^{2}}{2} - 16 u + 64 ln ∣ u ∣

Substitute back

u = x + 8

= \frac{( x + 8 ) ^{2}}{2} - 16 (x + 8) + 64 ln ∣ x + 8 ∣

Add a constant to the solution

= \frac{( x + 8 ) ^{2}}{2} - 16 (x + 8) + 64 ln ∣ x + 8 ∣ + C

\int \frac{1}{x 3 x} d x

\int_{1}^{8} \frac{x}{x ^{2} + 4 x + 13} d x

f (x) = 2 - 2 x

f^{^{''}} (x) = x^{2}, f^{^{'}} (0) = 6, f (0) = 3

x \to \infty lim (\frac{8 - e ^{x}}{8 + 3 e ^{x}})

What is the integral of (x^2)/(x+8) ?
The integral of (x^2)/(x+8) is ((x+8)^2)/2-16(x+8)+64ln|x+8|+C