What is the integral of (x^2+1)^22x ?

The integral of (x^2+1)^22x is (x^6)/3+x^4+x^2+C

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integral of (x^2+1)^22x

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

\frac{x ^{6}}{3} + x^{4} + x^{2} + C

Solution steps

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

Take the constant out:

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

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

Expand

(x^{2} + 1)^{2} x : x^{5} + 2 x^{3} + x

= 2 \cdot \int x^{5} + 2 x^{3} + x 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

= 2 (\int x^{5} d x + \int 2 x^{3} d x + \int x d x)

\int x^{5} d x = \frac{x ^{6}}{6}

\int 2 x^{3} d x = \frac{x ^{4}}{2}

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

= 2 (\frac{x ^{6}}{6} + \frac{x ^{4}}{2} + \frac{x ^{2}}{2})

Simplify

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

= \frac{x ^{6}}{3} + x^{4} + x^{2}

Add a constant to the solution

= \frac{x ^{6}}{3} + x^{4} + x^{2} + C

y^{^{'}} = \frac{t}{y e ^{- y^{2}} + y e ^{- y^{2}} t ^{2}}

\frac{d}{d x} (- \frac{14}{x ^{4}})

\int_{- \frac{π}{2}}^{\frac{π}{2}} tan (x) d x

\frac{s}{( s ^{2} + 1 ) ^{2}}

(\frac{( 3 x ^{2} )}{( 2 x + 4 )})^{3}

What is the integral of (x^2+1)^22x ?
The integral of (x^2+1)^22x is (x^6)/3+x^4+x^2+C