What is the integral of (sin(x)+yze^{xz}) ?

The integral of (sin(x)+yze^{xz}) is -cos(x)+ye^{zx}+C

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integral of (sin(x)+yze^{xz})

\int (sin (x) + yz e^{x z}) d x

- cos (x) + y e^{z x} + C

Solution steps

\int sin (x) + yz e^{x z} 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 yz e^{z x} d x

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

\int yz e^{z x} d x = y e^{z x}

= - cos (x) + y e^{z x}

Add a constant to the solution

= - cos (x) + y e^{z x} + C

\int \frac{2 x}{x ^{2} - 4 x + 5} d x

\int \frac{- 3}{x ( ln ( x ) ) ^{4}} d x

\int \frac{cos ( x )}{sin ^{2} ( 2 x )} d x

\int - 4 x^{4} e^{x^{5}} d x

What is the integral of (sin(x)+yze^{xz}) ?
The integral of (sin(x)+yze^{xz}) is -cos(x)+ye^{zx}+C