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

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

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

integral

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

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

Solution steps

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

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 = x + 4

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

Add a constant to the solution

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

y^{2} \cdot 2 (x^{2} - y^{2})^{\frac{1}{2}}

\frac{2 ln ( x )}{x}

\frac{1}{( y ^{2} + 6 y + 5 )}

x^{2} + 3 x

e^{x} (1 + x)

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