What is the integral of (x^2+1)^5*x ?

The integral of (x^2+1)^5*x is 1/12 (x^2+1)^6+C

Popular >

integral of (x^2+1)^5*x

\int (x^{2} + 1)^{5} \cdot x d x

\frac{1}{12} (x^{2} + 1)^{6} + C

Solution steps

\int (x^{2} + 1)^{5} x d x

Apply u-substitution

: \int \frac{u ^{5}}{2} d u

= \int \frac{u ^{5}}{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^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= \frac{1}{2} \cdot \frac{u ^{6}}{6}

Substitute back

u = x^{2} + 1

= \frac{1}{2} \cdot \frac{( x ^{2} + 1 ) ^{6}}{6}

Simplify

\frac{1}{2} \cdot \frac{( x ^{2} + 1 ) ^{6}}{6} : \frac{1}{12} (x^{2} + 1)^{6}

= \frac{1}{12} (x^{2} + 1)^{6}

Add a constant to the solution

= \frac{1}{12} (x^{2} + 1)^{6} + C

\frac{d ^{3}}{d x ^{3}} (ln (x))

\frac{d}{d x} (\frac{6}{x} - \frac{2}{x ^{3}} + \frac{1}{x ^{4}})

y = x^{- 8 c o s (x)}

y = sin (t) + \frac{1}{6} e^{t}, (π, \frac{1}{6} e^{π})

\frac{d}{d x} y = x^{4} \cdot x^{5}

What is the integral of (x^2+1)^5*x ?
The integral of (x^2+1)^5*x is 1/12 (x^2+1)^6+C