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Popular Calculus >

integral of (4x-1)^2

Solution

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

Solution

\frac{16 x ^{3}}{3} - 4 x^{2} + x + C

Solution steps

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

Expand

(4 x - 1)^{2} : 16 x^{2} - 8 x + 1

= \int 16 x^{2} - 8 x + 1 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 16 x^{2} d x - \int 8 x d x + \int 1 d x

\int 16 x^{2} d x = \frac{16 x ^{3}}{3}

\int 8 x d x = 4 x^{2}

\int 1 d x = x

= \frac{16 x ^{3}}{3} - 4 x^{2} + x

Add a constant to the solution

= \frac{16 x ^{3}}{3} - 4 x^{2} + x + C

Graph

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Frequently Asked Questions (FAQ)

What is the integral of (4x-1)^2 ?
The integral of (4x-1)^2 is (16x^3)/3-4x^2+x+C