What is the integral of (8x^3)/(x-5) ?

The integral of (8x^3)/(x-5) is 8((x^3)/3+(5x^2)/2+25x+125ln|x-5|)+C

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integral of (8x^3)/(x-5)

\int \frac{8 x ^{3}}{x - 5} d x

8 (\frac{x ^{3}}{3} + \frac{5 x ^{2}}{2} + 25 x + 125 ln ∣ x - 5 ∣) + C

Solution steps

\int \frac{8 x ^{3}}{x - 5} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int \frac{x ^{3}}{x - 5} d x

Long division

\frac{x ^{3}}{x - 5} : x^{2} + 5 x + 25 + \frac{125}{x - 5}

= 8 \cdot \int x^{2} + 5 x + 25 + \frac{125}{x - 5} 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

= 8 (\int x^{2} d x + \int 5 x d x + \int 25 d x + \int \frac{125}{x - 5} d x)

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

\int 5 x d x = \frac{5 x ^{2}}{2}

\int 25 d x = 25 x

\int \frac{125}{x - 5} d x = 125 ln ∣ x - 5 ∣

= 8 (\frac{x ^{3}}{3} + \frac{5 x ^{2}}{2} + 25 x + 125 ln ∣ x - 5 ∣)

Add a constant to the solution

= 8 (\frac{x ^{3}}{3} + \frac{5 x ^{2}}{2} + 25 x + 125 ln ∣ x - 5 ∣) + C

x \to 0 lim (\frac{( x + 2 ) ^{2} - 4}{x})

f (x) = x e^{x} - e^{x}

\int 7 e^{6} d x

t \to 1 lim (\frac{1}{t - 1})

x \to - 2 lim (\frac{\frac{4}{x} + 2}{x + 2})

What is the integral of (8x^3)/(x-5) ?
The integral of (8x^3)/(x-5) is 8((x^3)/3+(5x^2)/2+25x+125ln|x-5|)+C