What is the integral of (2ysin(x)+e^{2z}) ?

The integral of (2ysin(x)+e^{2z}) is e^{2z}y+y^2sin(x)+C

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integral of (2ysin(x)+e^{2z})

\int (2 y sin (x) + e^{2 z}) d y

e^{2 z} y + y^{2} sin (x) + C

Solution steps

\int 2 y sin (x) + e^{2 z} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int e^{2 z} d y + \int 2 sin (x) y d y

\int e^{2 z} d y = e^{2 z} y

\int 2 sin (x) y d y = y^{2} sin (x)

= e^{2 z} y + y^{2} sin (x)

Add a constant to the solution

= e^{2 z} y + y^{2} sin (x) + C

y^{^{''}} + 12 y^{^{'}} + 45 y = 0

\int_{1}^{2} \frac{9 x ^{3} + 4 x ^{2} - 2}{x} d x

\int \frac{sec ( x )}{tan ^{2} ( x )} d x

n = 1 \sum \infty 7 (3)^{n}

\int \frac{x}{d + x} d x

What is the integral of (2ysin(x)+e^{2z}) ?
The integral of (2ysin(x)+e^{2z}) is e^{2z}y+y^2sin(x)+C