What is the integral of (1-4x)(1+x) ?

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

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integral of (1-4x)(1+x)

integral

(1 - 4 x) (1 + x)

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

Solution steps

\int (1 - 4 x) (1 + x) d x

Expand

(1 - 4 x) (1 + x) : 1 - 3 x - 4 x^{2}

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

\int 1 d x = x

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

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

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

Add a constant to the solution

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

3 - 2 (x)^{- 1}

\frac{1}{( x )}

\frac{x ^{2}}{( 4 + x ^{6} )}

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

\frac{( 3 x - 13 )}{( x ^{2} - 12 + 45 ) ^{2}}

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