What is the integral of 5xe^{12x} ?

The integral of 5xe^{12x} is 5/144 (12e^{12x}x-e^{12x})+C

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integral of 5xe^{12x}

\int 5 x e^{12 x} d x

\frac{5}{144} (12 e^{12 x} x - e^{12 x}) + C

Solution steps

\int 5 x e^{12 x} d x

Take the constant out:

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

= 5 \cdot \int x e^{12 x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{144} d u

= 5 \cdot \int \frac{e ^{u} u}{144} d u

Take the constant out:

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

= 5 \cdot \frac{1}{144} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 5 \cdot \frac{1}{144} (e^{u} u - \int e^{u} d u)

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

= 5 \cdot \frac{1}{144} (e^{u} u - e^{u})

Substitute back

u = 12 x

= 5 \cdot \frac{1}{144} (e^{12 x} \cdot 12 x - e^{12 x})

Simplify

5 \cdot \frac{1}{144} (e^{12 x} \cdot 12 x - e^{12 x}) : \frac{5}{144} (12 e^{12 x} x - e^{12 x})

= \frac{5}{144} (12 e^{12 x} x - e^{12 x})

Add a constant to the solution

= \frac{5}{144} (12 e^{12 x} x - e^{12 x}) + C

y^{^{'}} + y = y^{- \frac{1}{2}}

\int_{- 2}^{1} \frac{1}{( x - 1 ) ^{3}} d x

f (x) = 1 - 2 x

\frac{d}{d x} (e^{- 3 x} sin (3 x))

y = x^{2} + x + 1

What is the integral of 5xe^{12x} ?
The integral of 5xe^{12x} is 5/144 (12e^{12x}x-e^{12x})+C