What is the integral of 3(2t+5)^3 ?

The integral of 3(2t+5)^3 is 3/8 (2t+5)^4+C

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integral of 3(2t+5)^3

\int 3 (2 t + 5)^{3} d t

\frac{3}{8} (2 t + 5)^{4} + C

Solution steps

\int 3 (2 t + 5)^{3} d t

Take the constant out:

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

= 3 \cdot \int (2 t + 5)^{3} d t

Apply u-substitution

: \int \frac{u ^{3}}{2} d u

= 3 \cdot \int \frac{u ^{3}}{2} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int u^{3} d u

Apply the Power Rule

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

= 3 \cdot \frac{1}{2} \cdot \frac{u ^{4}}{4}

Substitute back

u = 2 t + 5

= 3 \cdot \frac{1}{2} \cdot \frac{( 2 t + 5 ) ^{4}}{4}

Simplify

3 \cdot \frac{1}{2} \cdot \frac{( 2 t + 5 ) ^{4}}{4} : \frac{3}{8} (2 t + 5)^{4}

= \frac{3}{8} (2 t + 5)^{4}

Add a constant to the solution

= \frac{3}{8} (2 t + 5)^{4} + C

\frac{d}{d x} (arctan (ln (2 x + 1)))

\int \frac{( 82 x + 8 )}{( 9 x + 1 ) ( x - 1 )} d x

y^{^{'}} + 2 y = 4 t, y (0) = 3

x^{- 1}

\frac{d}{d x} (\frac{9 ( 5 ^{x} )}{x ^{5}})