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 ln (x^{2}) x d x

\int \frac{3 x}{x} d x

\int \frac{1 + a t ^{b}}{t ^{b}} d t

\int e^{x^{\frac{3}{2}}} d x

\int \frac{1}{c ^{2} + x ^{2}} d x

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