What is the integral of x(x^2+1)^4 ?

The integral of x(x^2+1)^4 is 1/10 (x^2+1)^5+C

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integral of x(x^2+1)^4

\int x (x^{2} + 1)^{4} d x

\frac{1}{10} (x^{2} + 1)^{5} + C

Solution steps

\int x (x^{2} + 1)^{4} d x

Apply u-substitution

: \int \frac{u ^{4}}{2} d u

= \int \frac{u ^{4}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= \frac{1}{2} \cdot \frac{u ^{5}}{5}

Substitute back

u = x^{2} + 1

= \frac{1}{2} \cdot \frac{( x ^{2} + 1 ) ^{5}}{5}

Simplify

\frac{1}{2} \cdot \frac{( x ^{2} + 1 ) ^{5}}{5} : \frac{1}{10} (x^{2} + 1)^{5}

= \frac{1}{10} (x^{2} + 1)^{5}

Add a constant to the solution

= \frac{1}{10} (x^{2} + 1)^{5} + C

e^{x} - 4 x^{3}

\frac{d}{d x} (\frac{x - 1}{x + 1})

\frac{\partial}{\partial x} (\frac{1}{2} ln (x + y))

y^{^{''}} + k^{2} y = 0

\frac{d}{d x} ((x + 2)^{2})

What is the integral of x(x^2+1)^4 ?
The integral of x(x^2+1)^4 is 1/10 (x^2+1)^5+C