What is the integral of (2xyz+sin(x)) ?

The integral of (2xyz+sin(x)) is -cos(x)+yzx^2+C

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integral of (2xyz+sin(x))

\int (2 x yz + sin (x)) d x

- cos (x) + yz x^{2} + C

Solution steps

\int 2 x yz + sin (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 sin (x) d x + \int 2 yz x d x

\int sin (x) d x = - cos (x)

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

= - cos (x) + yz x^{2}

Add a constant to the solution

= - cos (x) + yz x^{2} + C

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

\frac{d}{d x} (- \frac{5}{( x + 3 ) ^{3}})

\frac{1}{( 7 + x ) 2}

x \to 0 lim (x sin (x))

\int_{0}^{l n (21)} x e^{x} d x

What is the integral of (2xyz+sin(x)) ?
The integral of (2xyz+sin(x)) is -cos(x)+yzx^2+C