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

The integral of (4x+3)^4 is 1/20 (4x+3)^5+C

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integral of (4x+3)^4

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

\frac{1}{20} (4 x + 3)^{5} + C

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: \frac{u ^{5}}{5}

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

Substitute back

u = 4 x + 3

= \frac{1}{4} \cdot \frac{( 4 x + 3 ) ^{5}}{5}

Simplify

\frac{1}{4} \cdot \frac{( 4 x + 3 ) ^{5}}{5} : \frac{1}{20} (4 x + 3)^{5}

= \frac{1}{20} (4 x + 3)^{5}

Add a constant to the solution

= \frac{1}{20} (4 x + 3)^{5} + C

\int - 2 x sin (x) d x

f (x) = 6 x 4 - 3 x, at x = - 2

lo g_{2} (lo g_{2} (x))

(0, 4), (- 2, 0)