What is the integral of 32(8x+7)^3 ?

The integral of 32(8x+7)^3 is (8x+7)^4+C

integral of 32(8x+7)^3

\int 32 (8 x + 7)^{3} d x

(8 x + 7)^{4} + C

Solution steps

\int 32 (8 x + 7)^{3} d x

Take the constant out:

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

= 32 \cdot \int (8 x + 7)^{3} d x

Apply u-substitution

= 32 \cdot \int \frac{u ^{3}}{8} d u

Take the constant out:

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

= 32 \cdot \frac{1}{8} \cdot \int u^{3} d u

Apply the Power Rule

= 32 \cdot \frac{1}{8} \cdot \frac{u ^{4}}{4}

Substitute back

u = 8 x + 7

= 32 \cdot \frac{1}{8} \cdot \frac{( 8 x + 7 ) ^{4}}{4}

Simplify

32 \cdot \frac{1}{8} \cdot \frac{( 8 x + 7 ) ^{4}}{4} : (8 x + 7)^{4}

= (8 x + 7)^{4}

Add a constant to the solution

= (8 x + 7)^{4} + C

\int \frac{x ^{4} + 81}{x ( x ^{2} + 9 ) ^{2}} d x

in t e g r a l x^{- \frac{1}{2}}

\int x ln (1 + x^{2} + x) d x

\int x^{1.3} + 7 x^{2.5} d x

\int 5 cos (x) - 2 sec^{2} (x) d x