What is the integral of 4te^{4t} ?

The integral of 4te^{4t} is e^{4t}t-1/4 e^{4t}+C

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integral of 4te^{4t}

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

e^{4 t} t - \frac{1}{4} e^{4 t} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 4 \cdot \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

= 4 \cdot \frac{1}{16} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 4 \cdot \frac{1}{16} (e^{u} u - \int e^{u} d u)

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

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

Substitute back

u = 4 t

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

Simplify

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

= e^{4 t} t - \frac{1}{4} e^{4 t}

Add a constant to the solution

= e^{4 t} t - \frac{1}{4} e^{4 t} + C

\int 6 x^{2} + 4 d x

e^{\frac{t}{9}} sin (\frac{t}{9})

\int 8 (tan (8 x)) d x

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

f (x) = x^{2} + 1, (4, 17)

What is the integral of 4te^{4t} ?
The integral of 4te^{4t} is e^{4t}t-1/4 e^{4t}+C