What is the integral of (6x^4+7sqrt(x)+1/(x^3)) ?

The integral of (6x^4+7sqrt(x)+1/(x^3)) is (6x^5)/5+14/3 x^{3/2}-1/(2x^2)+C

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integral of (6x^4+7sqrt(x)+1/(x^3))

\int (6 x^{4} + 7 x + \frac{1}{x ^{3}}) d x

\frac{6 x ^{5}}{5} + \frac{14}{3} x^{\frac{3}{2}} - \frac{1}{2 x ^{2}} + C

Solution steps

\int 6 x^{4} + 7 x + \frac{1}{x ^{3}} 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 6 x^{4} d x + \int 7 x d x + \int \frac{1}{x ^{3}} d x

\int 6 x^{4} d x = \frac{6 x ^{5}}{5}

\int 7 x d x = \frac{14}{3} x^{\frac{3}{2}}

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

= \frac{6 x ^{5}}{5} + \frac{14}{3} x^{\frac{3}{2}} - \frac{1}{2 x ^{2}}

Add a constant to the solution

= \frac{6 x ^{5}}{5} + \frac{14}{3} x^{\frac{3}{2}} - \frac{1}{2 x ^{2}} + C

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What is the integral of (6x^4+7sqrt(x)+1/(x^3)) ?
The integral of (6x^4+7sqrt(x)+1/(x^3)) is (6x^5)/5+14/3 x^{3/2}-1/(2x^2)+C