What is the integral of 6x^3(3x^4+7)^7 ?

The integral of 6x^3(3x^4+7)^7 is

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integral of 6x^3(3x^4+7)^7

\int 6 x^{3} (3 x^{4} + 7)^{7} d x

6 (\frac{2187 x ^{32}}{32} + \frac{5103 x ^{28}}{4} + \frac{83349 x ^{24}}{8} + \frac{194481 x ^{20}}{4} + \frac{2268945 x ^{16}}{16} + \frac{1058841 x ^{12}}{4} + \frac{2470629 x ^{8}}{8} + \frac{823543 x ^{4}}{4}) + C

Solution steps

\int 6 x^{3} (3 x^{4} + 7)^{7} d x

Take the constant out:

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

= 6 \cdot \int x^{3} (3 x^{4} + 7)^{7} d x

Expand

x^{3} (3 x^{4} + 7)^{7} : 2187 x^{31} + 35721 x^{27} + 250047 x^{23} + 972405 x^{19} + 2268945 x^{15} + 3176523 x^{11} + 2470629 x^{7} + 823543 x^{3}

= 6 \cdot \int 2187 x^{31} + 35721 x^{27} + 250047 x^{23} + 972405 x^{19} + 2268945 x^{15} + 3176523 x^{11} + 2470629 x^{7} + 823543 x^{3} 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

= 6 (\int 2187 x^{31} d x + \int 35721 x^{27} d x + \int 250047 x^{23} d x + \int 972405 x^{19} d x + \int 2268945 x^{15} d x + \int 3176523 x^{11} d x + \int 2470629 x^{7} d x + \int 823543 x^{3} d x)

\int 2187 x^{31} d x = \frac{2187 x ^{32}}{32}

\int 35721 x^{27} d x = \frac{5103 x ^{28}}{4}

\int 250047 x^{23} d x = \frac{83349 x ^{24}}{8}

\int 972405 x^{19} d x = \frac{194481 x ^{20}}{4}

\int 2268945 x^{15} d x = \frac{2268945 x ^{16}}{16}

\int 3176523 x^{11} d x = \frac{1058841 x ^{12}}{4}

\int 2470629 x^{7} d x = \frac{2470629 x ^{8}}{8}

\int 823543 x^{3} d x = \frac{823543 x ^{4}}{4}

= 6 (\frac{2187 x ^{32}}{32} + \frac{5103 x ^{28}}{4} + \frac{83349 x ^{24}}{8} + \frac{194481 x ^{20}}{4} + \frac{2268945 x ^{16}}{16} + \frac{1058841 x ^{12}}{4} + \frac{2470629 x ^{8}}{8} + \frac{823543 x ^{4}}{4})

Add a constant to the solution

= 6 (\frac{2187 x ^{32}}{32} + \frac{5103 x ^{28}}{4} + \frac{83349 x ^{24}}{8} + \frac{194481 x ^{20}}{4} + \frac{2268945 x ^{16}}{16} + \frac{1058841 x ^{12}}{4} + \frac{2470629 x ^{8}}{8} + \frac{823543 x ^{4}}{4}) + C