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

The integral of (1+x^2)^3 is x+x^3+(3x^5)/5+(x^7)/7+C

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integral of (1+x^2)^3

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

x + x^{3} + \frac{3 x ^{5}}{5} + \frac{x ^{7}}{7} + C

Solution steps

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

Expand

(1 + x^{2})^{3} : 1 + 3 x^{2} + 3 x^{4} + x^{6}

= \int 1 + 3 x^{2} + 3 x^{4} + x^{6} 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

= \int 1 d x + \int 3 x^{2} d x + \int 3 x^{4} d x + \int x^{6} d x

\int 1 d x = x

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

\int 3 x^{4} d x = \frac{3 x ^{5}}{5}

\int x^{6} d x = \frac{x ^{7}}{7}

= x + x^{3} + \frac{3 x ^{5}}{5} + \frac{x ^{7}}{7}

Add a constant to the solution

= x + x^{3} + \frac{3 x ^{5}}{5} + \frac{x ^{7}}{7} + C

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What is the integral of (1+x^2)^3 ?
The integral of (1+x^2)^3 is x+x^3+(3x^5)/5+(x^7)/7+C