What is the integral of 5y*(2*ln(y)+e^{(4y^2)}) ?

The integral of 5y*(2*ln(y)+e^{(4y^2)}) is 5(y^2ln(y)-(y^2)/2+1/8 e^{4y^2})+C

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integral of 5y(2ln(y)+e^{(4y^2)})

\int 5 y \cdot (2 \cdot ln (y) + e^{(4 y^{2})}) d y

5 (y^{2} ln (y) - \frac{y ^{2}}{2} + \frac{1}{8} e^{4 y^{2}}) + C

Solution steps

\int 5 y (2 ln (y) + e^{4 y^{2}}) d y

Take the constant out:

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

= 5 \cdot \int y (2 ln (y) + e^{4 y^{2}}) d y

Expand

y (2 ln (y) + e^{4 y^{2}}) : 2 y ln (y) + e^{4 y^{2}} y

= 5 \cdot \int 2 y ln (y) + e^{4 y^{2}} y 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

= 5 (\int 2 y ln (y) d y + \int e^{4 y^{2}} y d y)

\int 2 y ln (y) d y = y^{2} ln (y) - \frac{y ^{2}}{2}

\int e^{4 y^{2}} y d y = \frac{1}{8} e^{4 y^{2}}

= 5 (y^{2} ln (y) - \frac{y ^{2}}{2} + \frac{1}{8} e^{4 y^{2}})

Add a constant to the solution

= 5 (y^{2} ln (y) - \frac{y ^{2}}{2} + \frac{1}{8} e^{4 y^{2}}) + C

What is the integral of 5y*(2*ln(y)+e^{(4y^2)}) ?
The integral of 5y*(2*ln(y)+e^{(4y^2)}) is 5(y^2ln(y)-(y^2)/2+1/8 e^{4y^2})+C