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

The integral of (x^4)/(x+4) is (x^4)/4-(4x^3)/3+8x^2-64x+256ln|x+4|+C

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

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

\frac{x ^{4}}{4} - \frac{4 x ^{3}}{3} + 8 x^{2} - 64 x + 256 ln ∣ x + 4 ∣ + C

Solution steps

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

Long division

\frac{x ^{4}}{x + 4} : x^{3} - 4 x^{2} + 16 x - 64 + \frac{256}{x + 4}

= \int x^{3} - 4 x^{2} + 16 x - 64 + \frac{256}{x + 4} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{3} d x - \int 4 x^{2} d x + \int 16 x d x - \int 64 d x + \int \frac{256}{x + 4} d x

\int x^{3} d x = \frac{x ^{4}}{4}

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int 16 x d x = 8 x^{2}

\int 64 d x = 64 x

\int \frac{256}{x + 4} d x = 256 ln ∣ x + 4 ∣

= \frac{x ^{4}}{4} - \frac{4 x ^{3}}{3} + 8 x^{2} - 64 x + 256 ln ∣ x + 4 ∣

Add a constant to the solution

= \frac{x ^{4}}{4} - \frac{4 x ^{3}}{3} + 8 x^{2} - 64 x + 256 ln ∣ x + 4 ∣ + C

\frac{\partial}{\partial z} (\frac{e ^{x z + y}}{z ^{2} + w})

sin (t)

\int ln (x^{3}) d x

\frac{d}{d x} (7 e^{x} (x))

\frac{x ^{4}}{4 - x ^{3}}

What is the integral of (x^4)/(x+4) ?
The integral of (x^4)/(x+4) is (x^4)/4-(4x^3)/3+8x^2-64x+256ln|x+4|+C