What is the integral of (x(6x^2+6))/(x^2+1) ?

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

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integral of (x(6x^2+6))/(x^2+1)

\int \frac{x ( 6 x ^{2} + 6 )}{x ^{2} + 1} d x

3 x^{2} + C

Solution steps

\int \frac{x ( 6 x ^{2} + 6 )}{x ^{2} + 1} d x

Long division

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

= \int 6 x d x

Take the constant out:

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

= 6 \cdot \int x d x

Apply the Power Rule

: \frac{x ^{2}}{2}

= 6 \cdot \frac{x ^{2}}{2}

Simplify

6 \cdot \frac{x ^{2}}{2} : 3 x^{2}

= 3 x^{2}

Add a constant to the solution

= 3 x^{2} + C

\frac{d}{d x} (- 2 sin (x) - sin (2 x))

\frac{d}{d x} ((x + 1) e^{a r c t a n (x)})

\frac{\partial}{\partial y} (x y^{2} + 5)

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

\int \frac{y ^{3}}{y ^{4} + 1} d y

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