What is the integral of (r^2)/(r+4) ?

The integral of (r^2)/(r+4) is ((r+4)^2)/2-8(r+4)+16ln|r+4|+C

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integral of (r^2)/(r+4)

\int \frac{r ^{2}}{r + 4} d r

\frac{( r + 4 ) ^{2}}{2} - 8 (r + 4) + 16 ln ∣ r + 4 ∣ + C

Solution steps

\int \frac{r ^{2}}{r + 4} d r

Apply u-substitution

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

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

Expand

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

= \int 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 u d u - \int 8 d u + \int \frac{16}{u} d u

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

\int 8 d u = 8 u

\int \frac{16}{u} d u = 16 ln ∣ u ∣

= \frac{u ^{2}}{2} - 8 u + 16 ln ∣ u ∣

Substitute back

u = r + 4

= \frac{( r + 4 ) ^{2}}{2} - 8 (r + 4) + 16 ln ∣ r + 4 ∣

Add a constant to the solution

= \frac{( r + 4 ) ^{2}}{2} - 8 (r + 4) + 16 ln ∣ r + 4 ∣ + C

\frac{x - 1}{x + 2}

\int \frac{x - 1}{4 x ^{2} - 8 x + 3} d x

\frac{\partial}{\partial x} (\frac{y}{2} x^{2})

\int_{0}^{\frac{π}{4}} (sec^{2} (x)) d x

\frac{d}{d x} (ln (sinh (x^{3})))

What is the integral of (r^2)/(r+4) ?
The integral of (r^2)/(r+4) is ((r+4)^2)/2-8(r+4)+16ln|r+4|+C