What is the integral of (x+4x-5i)^3 ?

The integral of (x+4x-5i)^3 is 1/4 (125x^4-750x^2)+i(125x-125x^3)+C

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integral of (x+4x-5i)^3

integral

(x + 4 x - 5 i)^{3}

\frac{1}{4} (125 x^{4} - 750 x^{2}) + i (125 x - 125 x^{3}) + C

Solution steps

\int (x + 4 x - 5 i)^{3} d x

Expand

(x + 4 x - 5 i)^{3} : (125 x^{3} - 375 x) + i (125 - 375 x^{2})

= \int 125 x^{3} - 375 x + i (125 - 375 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 125 x^{3} d x - \int 375 x d x + \int i (125 - 375 x^{2}) d x

\int 125 x^{3} d x = \frac{125 x ^{4}}{4}

\int 375 x d x = \frac{375 x ^{2}}{2}

\int i (125 - 375 x^{2}) d x = i (125 x - 125 x^{3})

= \frac{125 x ^{4}}{4} - \frac{375 x ^{2}}{2} + i (125 x - 125 x^{3})

Simplify

\frac{125 x ^{4}}{4} - \frac{375 x ^{2}}{2} + i (125 x - 125 x^{3}) : \frac{1}{4} (125 x^{4} - 750 x^{2}) + i (125 x - 125 x^{3})

= \frac{1}{4} (125 x^{4} - 750 x^{2}) + i (125 x - 125 x^{3})

Add a constant to the solution

= \frac{1}{4} (125 x^{4} - 750 x^{2}) + i (125 x - 125 x^{3}) + C

\frac{e ^{x}}{( e ) ^{2}} x + 1

\frac{1}{( 2 - y ^{2} )}

\frac{1}{( a + x ^{2} )}

\frac{1}{( 1 + sin ( x ) )}

x^{- 1} 0.01

What is the integral of (x+4x-5i)^3 ?
The integral of (x+4x-5i)^3 is 1/4 (125x^4-750x^2)+i(125x-125x^3)+C