What is the integral of (3+x^8)^48x^7 ?

The integral of (3+x^8)^48x^7 is 1/5 (3+x^8)^5+C

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integral of (3+x^8)^48x^7

\int (3 + x^{8})^{4} 8 x^{7} d x

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

Solution steps

\int (3 + x^{8})^{4} \cdot 8 x^{7} d x

Take the constant out:

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

= 8 \cdot \int (3 + x^{8})^{4} x^{7} d x

Apply u-substitution

: \int \frac{( 3 + u ) ^{4}}{8} d u

= 8 \cdot \int \frac{( 3 + u ) ^{4}}{8} d u

Take the constant out:

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

= 8 \cdot \frac{1}{8} \cdot \int (3 + u)^{4} d u

Apply u-substitution

: \int v^{4} d v

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

Apply the Power Rule

: \frac{v ^{5}}{5}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

f (x) = x^{2} - \frac{1}{x}, at x = - 1

\frac{d}{d x} (e^{a r c s i n (x)})

\frac{d}{d x} (x + 6 x + 3)

tan (x) + cos^{2} (x) + sin^{2} (x) - 3

\frac{d}{d x} ((x - 2)^{\frac{1}{2}})

What is the integral of (3+x^8)^48x^7 ?
The integral of (3+x^8)^48x^7 is 1/5 (3+x^8)^5+C