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

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

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

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

\frac{x ^{2}}{2} - \frac{4}{x} + C

Solution steps

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

Expand

\frac{x ^{3} + 4}{x ^{2}} : x + \frac{4}{x ^{2}}

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

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

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

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

Add a constant to the solution

= \frac{x ^{2}}{2} - \frac{4}{x} + C

\frac{\partial}{\partial x} (5 x^{5} y^{8} + 8 x^{6} y^{7})

\int_{0}^{\infty} \frac{x}{1 + x ^{4}} d x

\int e^{\frac{2 x}{π}} d x

\int \frac{( x + 1 )}{x ^{2} + 5 x + 6} d x

1 - 3 x

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