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

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

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integral of (3x^{1/2}+x^4)

\int (3 x^{\frac{1}{2}} + x^{4}) d x

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

Solution steps

\int 3 x^{\frac{1}{2}} + 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 3 x^{\frac{1}{2}} d x + \int x^{4} d x

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

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

= 2 x^{\frac{3}{2}} + \frac{x ^{5}}{5}

Add a constant to the solution

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

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