What is the integral of x^1(2+2x^2)^3 ?

The integral of x^1(2+2x^2)^3 is (8x^8)/7-(x^8)/7+(24x^6)/5-(4x^6)/5+6x^4+4x^2+C

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integral of x^1(2+2x^2)^3

\int x^{1} (2 + 2 x^{2})^{3} d x

\frac{8 x ^{8}}{7} - \frac{x ^{8}}{7} + \frac{24 x ^{6}}{5} - \frac{4 x ^{6}}{5} + 6 x^{4} + 4 x^{2} + C

Solution steps

\int x^{1} (2 + 2 x^{2})^{3} d x

Simplify

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

Apply Integration By Parts

: x (8 x + 8 x^{3} + \frac{24 x ^{5}}{5} + \frac{8 x ^{7}}{7}) - \int 8 x + 8 x^{3} + \frac{24 x ^{5}}{5} + \frac{8 x ^{7}}{7} d x

= x (8 x + 8 x^{3} + \frac{24 x ^{5}}{5} + \frac{8 x ^{7}}{7}) - \int 8 x + 8 x^{3} + \frac{24 x ^{5}}{5} + \frac{8 x ^{7}}{7} d x

\int 8 x + 8 x^{3} + \frac{24 x ^{5}}{5} + \frac{8 x ^{7}}{7} d x = 4 x^{2} + 2 x^{4} + \frac{4 x ^{6}}{5} + \frac{x ^{8}}{7}

= x (8 x + 8 x^{3} + \frac{24 x ^{5}}{5} + \frac{8 x ^{7}}{7}) - (4 x^{2} + 2 x^{4} + \frac{4 x ^{6}}{5} + \frac{x ^{8}}{7})

Simplify

x (8 x + 8 x^{3} + \frac{24 x ^{5}}{5} + \frac{8 x ^{7}}{7}) - (4 x^{2} + 2 x^{4} + \frac{4 x ^{6}}{5} + \frac{x ^{8}}{7}) : \frac{8 x ^{8}}{7} - \frac{x ^{8}}{7} + \frac{24 x ^{6}}{5} - \frac{4 x ^{6}}{5} + 6 x^{4} + 4 x^{2}

= \frac{8 x ^{8}}{7} - \frac{x ^{8}}{7} + \frac{24 x ^{6}}{5} - \frac{4 x ^{6}}{5} + 6 x^{4} + 4 x^{2}

Add a constant to the solution

= \frac{8 x ^{8}}{7} - \frac{x ^{8}}{7} + \frac{24 x ^{6}}{5} - \frac{4 x ^{6}}{5} + 6 x^{4} + 4 x^{2} + C

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What is the integral of x^1(2+2x^2)^3 ?
The integral of x^1(2+2x^2)^3 is (8x^8)/7-(x^8)/7+(24x^6)/5-(4x^6)/5+6x^4+4x^2+C