What is the integral of t*e^{4t} ?

The integral of t*e^{4t} is 1/16 (e^{4t}*4t-e^{4t})+C

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integral of t*e^{4t}

\int t \cdot e^{4 t} d t

\frac{1}{16} (e^{4 t} \cdot 4 t - e^{4 t}) + C

Solution steps

\int t e^{4 t} d t

Apply u-substitution

: \int \frac{e ^{u} u}{16} d u

= \int \frac{e ^{u} u}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{16} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{16} (e^{u} u - e^{u})

Substitute back

u = 4 t

= \frac{1}{16} (e^{4 t} \cdot 4 t - e^{4 t})

Add a constant to the solution

= \frac{1}{16} (e^{4 t} \cdot 4 t - e^{4 t}) + C

f (x) = \frac{1 + x + x ^{2}}{1 - x ^{2}}

f (x) = \frac{4 x + 5}{4 x - 5}

\frac{d}{d x} (ln (sin (x)) + cos^{2} (x))

\int \frac{y ^{2} - 81}{y} d y

n = 0 \sum \infty \frac{1}{2 ^{n} + 3 ^{n}}

What is the integral of t*e^{4t} ?
The integral of t*e^{4t} is 1/16 (e^{4t}*4t-e^{4t})+C