What is the integral of (8x+3)^4 ?

The integral of (8x+3)^4 is 1/40 (8x+3)^5+C

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integral of (8x+3)^4

\int (8 x + 3)^{4} d x

\frac{1}{40} (8 x + 3)^{5} + C

Solution steps

\int (8 x + 3)^{4} d x

Apply u-substitution

: \int \frac{u ^{4}}{8} d u

= \int \frac{u ^{4}}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int u^{4} d u

Apply the Power Rule

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

= \frac{1}{8} \cdot \frac{u ^{5}}{5}

Substitute back

u = 8 x + 3

= \frac{1}{8} \cdot \frac{( 8 x + 3 ) ^{5}}{5}

Simplify

\frac{1}{8} \cdot \frac{( 8 x + 3 ) ^{5}}{5} : \frac{1}{40} (8 x + 3)^{5}

= \frac{1}{40} (8 x + 3)^{5}

Add a constant to the solution

= \frac{1}{40} (8 x + 3)^{5} + C

\frac{d ^{2}}{d x ^{2}} (e^{x} - x^{4})

(48 - 2 x) - x, [- 8, 24]

x \to \infty lim (\frac{- 5}{x})

\frac{1}{( s - 3 )}

y^{^{'}} = x 1 - y^{2}