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

The integral of (4x-3)^5 is 1/24 (4x-3)^6+C

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integral of (4x-3)^5

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

\frac{1}{24} (4 x - 3)^{6} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{5}}{4} d u

= \int \frac{u ^{5}}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= \frac{1}{4} \cdot \frac{u ^{6}}{6}

Substitute back

u = 4 x - 3

= \frac{1}{4} \cdot \frac{( 4 x - 3 ) ^{6}}{6}

Simplify

\frac{1}{4} \cdot \frac{( 4 x - 3 ) ^{6}}{6} : \frac{1}{24} (4 x - 3)^{6}

= \frac{1}{24} (4 x - 3)^{6}

Add a constant to the solution

= \frac{1}{24} (4 x - 3)^{6} + C

\int cos (n wt) d t

y^{^{''}} + 9 y = 3 sin (3 t)

y = (1 - e^{x})^{4}

\frac{\partial}{\partial x} (\frac{1}{2} m v^{2})

y (θ) = 2 - θ^{5} sin (θ)