What is the integral of 3x^2(x^3+7)^3 ?

The integral of 3x^2(x^3+7)^3 is (x^{12})/4+7x^9+(147x^6)/2+343x^3+C

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integral of 3x^2(x^3+7)^3

\int 3 x^{2} (x^{3} + 7)^{3} d x

\frac{x ^{12}}{4} + 7 x^{9} + \frac{147 x ^{6}}{2} + 343 x^{3} + C

Solution steps

\int 3 x^{2} (x^{3} + 7)^{3} d x

Take the constant out:

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

= 3 \cdot \int x^{2} (x^{3} + 7)^{3} d x

Expand

x^{2} (x^{3} + 7)^{3} : x^{11} + 21 x^{8} + 147 x^{5} + 343 x^{2}

= 3 \cdot \int x^{11} + 21 x^{8} + 147 x^{5} + 343 x^{2} d x

Apply the Sum Rule:

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

= 3 (\int x^{11} d x + \int 21 x^{8} d x + \int 147 x^{5} d x + \int 343 x^{2} d x)

\int x^{11} d x = \frac{x ^{12}}{12}

\int 21 x^{8} d x = \frac{7 x ^{9}}{3}

\int 147 x^{5} d x = \frac{49 x ^{6}}{2}

\int 343 x^{2} d x = \frac{343 x ^{3}}{3}

= 3 (\frac{x ^{12}}{12} + \frac{7 x ^{9}}{3} + \frac{49 x ^{6}}{2} + \frac{343 x ^{3}}{3})

Simplify

3 (\frac{x ^{12}}{12} + \frac{7 x ^{9}}{3} + \frac{49 x ^{6}}{2} + \frac{343 x ^{3}}{3}) : \frac{x ^{12}}{4} + 7 x^{9} + \frac{147 x ^{6}}{2} + 343 x^{3}

= \frac{x ^{12}}{4} + 7 x^{9} + \frac{147 x ^{6}}{2} + 343 x^{3}

Add a constant to the solution

= \frac{x ^{12}}{4} + 7 x^{9} + \frac{147 x ^{6}}{2} + 343 x^{3} + C

(9 x y^{2} - 3) d x + (9 x^{2} y + 4) d y = 0

f (x) = e^{x + 9} + 9

\int 3 x 7 - 3 x^{2} d x

\int \frac{1}{x ( 1 + ln ( x ) )} d x

t^{2} y^{^{''}} + t y^{^{'}} + y = 0

What is the integral of 3x^2(x^3+7)^3 ?
The integral of 3x^2(x^3+7)^3 is (x^{12})/4+7x^9+(147x^6)/2+343x^3+C