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Popular Calculus >

integral of (4z+1)^2

Solution

\int (4 z + 1)^{2} d z

Solution

\frac{16 z ^{3}}{3} + 4 z^{2} + z + C

Solution steps

\int (4 z + 1)^{2} d z

Expand

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

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

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 z^{2} d z + \int 8 z d z + \int 1 d z

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

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

\int 1 d z = z

= \frac{16 z ^{3}}{3} + 4 z^{2} + z

Add a constant to the solution

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

Graph

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(\partial)/(\partial x)(4^x)

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Frequently Asked Questions (FAQ)

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