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

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

integral of y(y^2+1)^5

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

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

Solution steps

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

Apply u-substitution

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

Substitute back

u = y^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

x \to - 6 lim (\frac{( x + 7 )}{x + 6})

\int \frac{6 e ^{2 x}}{e ^{2 x} + 12 e ^{x} + 27} d x

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

\int x (a - b x^{2}) d x

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

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