What is the integral of (x^5-3125)/(x-5) ?

The integral of (x^5-3125)/(x-5) is (x^5)/5+(5x^4)/4+(25x^3)/3+(125x^2)/2+625x+C

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integral of (x^5-3125)/(x-5)

\int \frac{x ^{5} - 3125}{x - 5} d x

\frac{x ^{5}}{5} + \frac{5 x ^{4}}{4} + \frac{25 x ^{3}}{3} + \frac{125 x ^{2}}{2} + 625 x + C

Solution steps

\int \frac{x ^{5} - 3125}{x - 5} d x

Simplify

\frac{x ^{5} - 3125}{x - 5} : x^{4} + 5 x^{3} + 25 x^{2} + 125 x + 625

= \int x^{4} + 5 x^{3} + 25 x^{2} + 125 x + 625 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^{4} d x + \int 5 x^{3} d x + \int 25 x^{2} d x + \int 125 x d x + \int 625 d x

\int x^{4} d x = \frac{x ^{5}}{5}

\int 5 x^{3} d x = \frac{5 x ^{4}}{4}

\int 25 x^{2} d x = \frac{25 x ^{3}}{3}

\int 125 x d x = \frac{125 x ^{2}}{2}

\int 625 d x = 625 x

= \frac{x ^{5}}{5} + \frac{5 x ^{4}}{4} + \frac{25 x ^{3}}{3} + \frac{125 x ^{2}}{2} + 625 x

Add a constant to the solution

= \frac{x ^{5}}{5} + \frac{5 x ^{4}}{4} + \frac{25 x ^{3}}{3} + \frac{125 x ^{2}}{2} + 625 x + C

What is the integral of (x^5-3125)/(x-5) ?
The integral of (x^5-3125)/(x-5) is (x^5)/5+(5x^4)/4+(25x^3)/3+(125x^2)/2+625x+C