What is the integral of (5x^5+1)/(x^4) ?

The integral of (5x^5+1)/(x^4) is (5x^2)/2-1/(3x^3)+C

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integral of (5x^5+1)/(x^4)

\int \frac{5 x ^{5} + 1}{x ^{4}} d x

\frac{5 x ^{2}}{2} - \frac{1}{3 x ^{3}} + C

Solution steps

\int \frac{5 x ^{5} + 1}{x ^{4}} d x

Expand

\frac{5 x ^{5} + 1}{x ^{4}} : 5 x + \frac{1}{x ^{4}}

= \int 5 x + \frac{1}{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 5 x d x + \int \frac{1}{x ^{4}} d x

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

\int \frac{1}{x ^{4}} d x = - \frac{1}{3 x ^{3}}

= \frac{5 x ^{2}}{2} - \frac{1}{3 x ^{3}}

Add a constant to the solution

= \frac{5 x ^{2}}{2} - \frac{1}{3 x ^{3}} + C

\frac{s}{5}

16 e^{x}

\int t e^{6 t} d t

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

sin (y)

What is the integral of (5x^5+1)/(x^4) ?
The integral of (5x^5+1)/(x^4) is (5x^2)/2-1/(3x^3)+C