What is the integral of (3x^2+9/(2x^3)) ?

The integral of (3x^2+9/(2x^3)) is x^3-9/(4x^2)+C

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integral of (3x^2+9/(2x^3))

\int (3 x^{2} + \frac{9}{2 x ^{3}}) d x

x^{3} - \frac{9}{4 x ^{2}} + C

Solution steps

\int 3 x^{2} + \frac{9}{2 x ^{3}} 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^{2} d x + \int \frac{9}{2 x ^{3}} d x

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

\int \frac{9}{2 x ^{3}} d x = - \frac{9}{4 x ^{2}}

= x^{3} - \frac{9}{4 x ^{2}}

Add a constant to the solution

= x^{3} - \frac{9}{4 x ^{2}} + C

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

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

\int 24 (1 + x^{2}) (7 + 12 x + 4 x^{3})^{10} d x

\int \frac{7}{x + 4} d x

What is the integral of (3x^2+9/(2x^3)) ?
The integral of (3x^2+9/(2x^3)) is x^3-9/(4x^2)+C