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

The integral of (3/(x^2)+4-1/(4x^3)) is -3/x+4x+1/(8x^2)+C

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

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

- \frac{3}{x} + 4 x + \frac{1}{8 x ^{2}} + C

Solution steps

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

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

\int 4 d x = 4 x

\int \frac{1}{4 x ^{3}} d x = - \frac{1}{8 x ^{2}}

= - \frac{3}{x} + 4 x - (- \frac{1}{8 x ^{2}})

Simplify

= - \frac{3}{x} + 4 x + \frac{1}{8 x ^{2}}

Add a constant to the solution

= - \frac{3}{x} + 4 x + \frac{1}{8 x ^{2}} + C

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What is the integral of (3/(x^2)+4-1/(4x^3)) ?
The integral of (3/(x^2)+4-1/(4x^3)) is -3/x+4x+1/(8x^2)+C