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

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

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

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

\frac{x ^{8}}{8} + \frac{x ^{6}}{2} + \frac{3 x ^{4}}{4} + \frac{x ^{2}}{2} + C

Solution steps

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

Expand

(x^{2} + 1)^{3} x : x^{7} + 3 x^{5} + 3 x^{3} + x

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

= \int x^{7} d x + \int 3 x^{5} d x + \int 3 x^{3} d x + \int x d x

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

\int 3 x^{5} d x = \frac{x ^{6}}{2}

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

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

= \frac{x ^{8}}{8} + \frac{x ^{6}}{2} + \frac{3 x ^{4}}{4} + \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{8}}{8} + \frac{x ^{6}}{2} + \frac{3 x ^{4}}{4} + \frac{x ^{2}}{2} + C

\frac{d}{d x} (\frac{4 x ^{2} + 6 x + 8}{x})

\int \frac{y}{1 + x ^{2}} d y

f (x) = 4 + 4 x^{2} - 2 x^{3}, at x = a

t \to 0 lim (sin (6 t))

d y - (2 y - e^{2 x}) d x = 0

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