What is the integral of 8xe^{15x} ?

The integral of 8xe^{15x} is 8/225 (15e^{15x}x-e^{15x})+C

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integral of 8xe^{15x}

\int 8 x e^{15 x} d x

\frac{8}{225} (15 e^{15 x} x - e^{15 x}) + C

Solution steps

\int 8 x e^{15 x} d x

Take the constant out:

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

= 8 \cdot \int x e^{15 x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{225} d u

= 8 \cdot \int \frac{e ^{u} u}{225} d u

Take the constant out:

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

= 8 \cdot \frac{1}{225} \cdot \int e^{u} u d u

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

= 8 \cdot \frac{1}{225} (e^{u} u - \int e^{u} d u)

\int e^{u} d u = e^{u}

= 8 \cdot \frac{1}{225} (e^{u} u - e^{u})

Substitute back

u = 15 x

= 8 \cdot \frac{1}{225} (e^{15 x} \cdot 15 x - e^{15 x})

Simplify

8 \cdot \frac{1}{225} (e^{15 x} \cdot 15 x - e^{15 x}) : \frac{8}{225} (15 e^{15 x} x - e^{15 x})

= \frac{8}{225} (15 e^{15 x} x - e^{15 x})

Add a constant to the solution

= \frac{8}{225} (15 e^{15 x} x - e^{15 x}) + C

\frac{d}{d x} (\frac{- x ^{2} + y ^{2}}{( x ^{2} + y ^{2} ) ^{2}})

\frac{d y}{d t} = - \frac{y}{t} + 2

\frac{d y}{d x}, x y + x = 2

\int \frac{a ^{2} + x ^{2}}{x ^{2}} d x

y = 6 x - x^{2}, y = x

What is the integral of 8xe^{15x} ?
The integral of 8xe^{15x} is 8/225 (15e^{15x}x-e^{15x})+C