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

The integral of (x^5+1/(x^3)+4) is (x^6)/6-1/(2x^2)+4x+C

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integral of (x^5+1/(x^3)+4)

\int (x^{5} + \frac{1}{x ^{3}} + 4) d x

\frac{x ^{6}}{6} - \frac{1}{2 x ^{2}} + 4 x + C

Solution steps

\int x^{5} + \frac{1}{x ^{3}} + 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^{5} d x + \int \frac{1}{x ^{3}} d x + \int 4 d x

\int x^{5} d x = \frac{x ^{6}}{6}

\int \frac{1}{x ^{3}} d x = - \frac{1}{2 x ^{2}}

\int 4 d x = 4 x

= \frac{x ^{6}}{6} - \frac{1}{2 x ^{2}} + 4 x

Add a constant to the solution

= \frac{x ^{6}}{6} - \frac{1}{2 x ^{2}} + 4 x + C

\int - 3 x^{4} e^{x^{5}} d x

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

\int \frac{x}{( x ^{2} + 4 ) ^{15}} d x

\int arccos (\frac{x}{2}) d x

What is the integral of (x^5+1/(x^3)+4) ?
The integral of (x^5+1/(x^3)+4) is (x^6)/6-1/(2x^2)+4x+C