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

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

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integral of 4x^3-3/(x^4)

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

x^{4} + \frac{1}{x ^{3}} + C

Solution steps

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

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

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

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

Simplify

= x^{4} + \frac{1}{x ^{3}}

Add a constant to the solution

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

3 - 9 x

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

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

\int \frac{1}{2 x} cos (2 x) d x

x \to 8 - lim (\frac{x - 9}{x - 8})

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