What is the integral of (7+x^5)^65x^4 ?

The integral of (7+x^5)^65x^4 is 1/7 (7+x^5)^7+C

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integral of (7+x^5)^65x^4

\int (7 + x^{5})^{6} 5 x^{4} d x

\frac{1}{7} (7 + x^{5})^{7} + C

Solution steps

\int (7 + x^{5})^{6} \cdot 5 x^{4} d x

Take the constant out:

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

= 5 \cdot \int (7 + x^{5})^{6} x^{4} d x

Apply u-substitution

: \int \frac{u ^{6}}{5} d u

= 5 \cdot \int \frac{u ^{6}}{5} d u

Take the constant out:

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

= 5 \cdot \frac{1}{5} \cdot \int u^{6} d u

Apply the Power Rule

: \frac{u ^{7}}{7}

= 5 \cdot \frac{1}{5} \cdot \frac{u ^{7}}{7}

Substitute back

u = 7 + x^{5}

= 5 \cdot \frac{1}{5} \cdot \frac{( 7 + x ^{5} ) ^{7}}{7}

Simplify

5 \cdot \frac{1}{5} \cdot \frac{( 7 + x ^{5} ) ^{7}}{7} : \frac{1}{7} (7 + x^{5})^{7}

= \frac{1}{7} (7 + x^{5})^{7}

Add a constant to the solution

= \frac{1}{7} (7 + x^{5})^{7} + C

\frac{d}{d x} (cos (\frac{1}{2} x))

y^{^{''}} - 16 y^{^{'}} + 64 y = 8 t^{6} e^{8 t}

\frac{d}{d x} (x^{2} + 17)

\frac{d}{d x} (2 ln (x) + 3)

\int (x^{20} - 5 x^{4})^{5} (x^{19} - x^{3}) d x

What is the integral of (7+x^5)^65x^4 ?
The integral of (7+x^5)^65x^4 is 1/7 (7+x^5)^7+C