What is the integral of 1/3 x^3(x^2+1)^{65} ?

The integral of 1/3 x^3(x^2+1)^{65} is 1/6 (1/67 (x^2+1)^{67}-1/66 (x^2+1)^{66})+C

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integral of 1/3 x^3(x^2+1)^{65}

\int \frac{1}{3} x^{3} (x^{2} + 1)^{65} d x

\frac{1}{6} (\frac{1}{67} (x^{2} + 1)^{67} - \frac{1}{66} (x^{2} + 1)^{66}) + C

Solution steps

\int \frac{1}{3} x^{3} (x^{2} + 1)^{65} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int x^{3} (x^{2} + 1)^{65} d x

Apply u-substitution

: \int \frac{( u - 1 ) u ^{65}}{2} d u

= \frac{1}{3} \cdot \int \frac{( u - 1 ) u ^{65}}{2} d u

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{2} \cdot \int (u - 1) u^{65} d u

Expand

(u - 1) u^{65} : u^{66} - u^{65}

= \frac{1}{3} \cdot \frac{1}{2} \cdot \int u^{66} - u^{65} d u

Apply the Sum Rule:

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

= \frac{1}{3} \cdot \frac{1}{2} (\int u^{66} d u - \int u^{65} d u)

\int u^{66} d u = \frac{u ^{67}}{67}

\int u^{65} d u = \frac{u ^{66}}{66}

= \frac{1}{3} \cdot \frac{1}{2} (\frac{u ^{67}}{67} - \frac{u ^{66}}{66})

Substitute back

u = x^{2} + 1

= \frac{1}{3} \cdot \frac{1}{2} (\frac{( x ^{2} + 1 ) ^{67}}{67} - \frac{( x ^{2} + 1 ) ^{66}}{66})

Simplify

\frac{1}{3} \cdot \frac{1}{2} (\frac{( x ^{2} + 1 ) ^{67}}{67} - \frac{( x ^{2} + 1 ) ^{66}}{66}) : \frac{1}{6} (\frac{1}{67} (x^{2} + 1)^{67} - \frac{1}{66} (x^{2} + 1)^{66})

= \frac{1}{6} (\frac{1}{67} (x^{2} + 1)^{67} - \frac{1}{66} (x^{2} + 1)^{66})

Add a constant to the solution

= \frac{1}{6} (\frac{1}{67} (x^{2} + 1)^{67} - \frac{1}{66} (x^{2} + 1)^{66}) + C

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What is the integral of 1/3 x^3(x^2+1)^{65} ?
The integral of 1/3 x^3(x^2+1)^{65} is 1/6 (1/67 (x^2+1)^{67}-1/66 (x^2+1)^{66})+C