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

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

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

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

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

Solution steps

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

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (19)

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