What is the integral of (12x+3x^4)(10x^2+x^5-2)^6 ?

The integral of (12x+3x^4)(10x^2+x^5-2)^6 is 3/35 (10x^2+x^5-2)^7+C

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integral of (12x+3x^4)(10x^2+x^5-2)^6

\int (12 x + 3 x^{4}) (10 x^{2} + x^{5} - 2)^{6} d x

\frac{3}{35} (10 x^{2} + x^{5} - 2)^{7} + C

Solution steps

\int (12 x + 3 x^{4}) (10 x^{2} + x^{5} - 2)^{6} d x

Apply u-substitution

: \int \frac{3 u ^{6}}{5} d u

= \int \frac{3 u ^{6}}{5} d u

Take the constant out:

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

= \frac{3}{5} \cdot \int u^{6} d u

Apply the Power Rule

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

= \frac{3}{5} \cdot \frac{u ^{7}}{7}

Substitute back

u = 10 x^{2} + x^{5} - 2

= \frac{3}{5} \cdot \frac{( 10 x ^{2} + x ^{5} - 2 ) ^{7}}{7}

Simplify

\frac{3}{5} \cdot \frac{( 10 x ^{2} + x ^{5} - 2 ) ^{7}}{7} : \frac{3}{35} (10 x^{2} + x^{5} - 2)^{7}

= \frac{3}{35} (10 x^{2} + x^{5} - 2)^{7}

Add a constant to the solution

= \frac{3}{35} (10 x^{2} + x^{5} - 2)^{7} + C

x \to 0 lim (3 x^{2} + x)

\frac{d}{d x} (\frac{8 - sec ( x )}{tan ( x )})

\frac{\partial}{\partial z} (x e^{yz})

x^{2} + 4 x + 2

x \to π lim (π)

What is the integral of (12x+3x^4)(10x^2+x^5-2)^6 ?
The integral of (12x+3x^4)(10x^2+x^5-2)^6 is 3/35 (10x^2+x^5-2)^7+C