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

The integral of ((x^3)/3+1/(4x^2)) is (x^4)/(12)-1/(4x)+C

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

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

\frac{x ^{4}}{12} - \frac{1}{4 x} + C

Solution steps

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

\int \frac{x ^{3}}{3} d x = \frac{x ^{4}}{12}

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

= \frac{x ^{4}}{12} - \frac{1}{4 x}

Add a constant to the solution

= \frac{x ^{4}}{12} - \frac{1}{4 x} + C

What is the integral of ((x^3)/3+1/(4x^2)) ?
The integral of ((x^3)/3+1/(4x^2)) is (x^4)/(12)-1/(4x)+C