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

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

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

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

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

Solution steps

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

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

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

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

Add a constant to the solution

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

\int (\frac{u ^{3} + 2 u ^{2} - u}{3 u}) d u

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

\int (5^{y} + \frac{6 x}{y + 4}) d x

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