What is the integral of (3x^4-2x^{-3}+sec^2(x)) ?

The integral of (3x^4-2x^{-3}+sec^2(x)) is (3x^5)/5+1/(x^2)+tan(x)+C

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integral of (3x^4-2x^{-3}+sec^2(x))

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

\frac{3 x ^{5}}{5} + \frac{1}{x ^{2}} + tan (x) + C

Solution steps

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

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

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

\int 3 x^{4} d x = \frac{3 x ^{5}}{5}

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

\int sec^{2} (x) d x = tan (x)

= \frac{3 x ^{5}}{5} - (- \frac{1}{x ^{2}}) + tan (x)

Simplify

= \frac{3 x ^{5}}{5} + \frac{1}{x ^{2}} + tan (x)

Add a constant to the solution

= \frac{3 x ^{5}}{5} + \frac{1}{x ^{2}} + tan (x) + C

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What is the integral of (3x^4-2x^{-3}+sec^2(x)) ?
The integral of (3x^4-2x^{-3}+sec^2(x)) is (3x^5)/5+1/(x^2)+tan(x)+C