What is the integral of (6+x^5)^25x^4 ?

The integral of (6+x^5)^25x^4 is 36x^5+6x^{10}+(x^{15})/3+C

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integral of (6+x^5)^25x^4

\int (6 + x^{5})^{2} 5 x^{4} d x

36 x^{5} + 6 x^{10} + \frac{x ^{15}}{3} + C

Solution steps

\int (6 + x^{5})^{2} \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 (6 + x^{5})^{2} x^{4} d x

Apply u-substitution

: \int \frac{( 6 + u ) ^{2}}{5} d u

= 5 \cdot \int \frac{( 6 + u ) ^{2}}{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 (6 + u)^{2} d u

Expand

(6 + u)^{2} : 36 + 12 u + u^{2}

= 5 \cdot \frac{1}{5} \cdot \int 36 + 12 u + u^{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 5 \cdot \frac{1}{5} (\int 36 d u + \int 12 u d u + \int u^{2} d u)

\int 36 d u = 36 u

\int 12 u d u = 6 u^{2}

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

= 5 \cdot \frac{1}{5} (36 u + 6 u^{2} + \frac{u ^{3}}{3})

Substitute back

u = x^{5}

= 5 \cdot \frac{1}{5} (36 x^{5} + 6 (x^{5})^{2} + \frac{( x ^{5} ) ^{3}}{3})

Simplify

5 \cdot \frac{1}{5} (36 x^{5} + 6 (x^{5})^{2} + \frac{( x ^{5} ) ^{3}}{3}) : 36 x^{5} + 6 x^{10} + \frac{x ^{15}}{3}

= 36 x^{5} + 6 x^{10} + \frac{x ^{15}}{3}

Add a constant to the solution

= 36 x^{5} + 6 x^{10} + \frac{x ^{15}}{3} + C

\frac{\partial}{\partial x} (- cos (2 x))

\frac{\partial}{\partial x} (x^{3} + y^{3} - 3 x^{2} + 3 y^{2})

\frac{d}{d a} (1 + cos^{4} (a) - sin^{4} (a))

x \to \infty lim (\frac{2 x}{x})

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{3 x} sin (2 x)

What is the integral of (6+x^5)^25x^4 ?
The integral of (6+x^5)^25x^4 is 36x^5+6x^{10}+(x^{15})/3+C