What is the integral of t2^{t+2} ?

The integral of t2^{t+2} is 4((2^tt)/(ln(2))-(2^t)/(ln^2(2)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of t2^{t+2}

\int t 2^{t + 2} d t

4 (\frac{2 ^{t} t}{ln ( 2 )} - \frac{2 ^{t}}{ln ^{2} ( 2 )}) + C

Solution steps

\int t \cdot 2^{t + 2} d t

t 2^{t + 2} = 2^{2} t 2^{t}

= \int 2^{2} t \cdot 2^{t} d t

Take the constant out:

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

= 2^{2} \cdot \int t \cdot 2^{t} d t

Simplify

= 4 \cdot \int t \cdot 2^{t} d t

Apply Integration By Parts

= 4 (\frac{2 ^{t} t}{ln ( 2 )} - \int \frac{2 ^{t}}{ln ( 2 )} d t)

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

= 4 (\frac{2 ^{t} t}{ln ( 2 )} - \frac{2 ^{t}}{ln ^{2} ( 2 )})

Add a constant to the solution

= 4 (\frac{2 ^{t} t}{ln ( 2 )} - \frac{2 ^{t}}{ln ^{2} ( 2 )}) + C

What is the integral of t2^{t+2} ?
The integral of t2^{t+2} is 4((2^tt)/(ln(2))-(2^t)/(ln^2(2)))+C