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

The integral of t2^{t^2} is (2^{t^2-1})/(ln(2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of t2^{t^2}

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

\frac{2 ^{t^{2} - 1}}{ln ( 2 )} + C

Solution steps

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

Apply u-substitution

: \int 2^{u - 1} d u

= \int 2^{u - 1} d u

Simplify

2^{u - 1} : 2^{u} \cdot 2^{- 1}

= \int 2^{u} \cdot 2^{- 1} d u

Take the constant out:

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

= 2^{- 1} \cdot \int 2^{u} d u

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \frac{1}{2} \cdot \int 2^{u} d u

\int a^{x} d x = \frac{a ^{x}}{ln a}

= \frac{1}{2} \cdot \frac{2 ^{u}}{ln ( 2 )}

Substitute back

u = t^{2}

= \frac{1}{2} \cdot \frac{2 ^{t^{2}}}{ln ( 2 )}

Simplify

\frac{1}{2} \cdot \frac{2 ^{t^{2}}}{ln ( 2 )} : \frac{2 ^{t^{2} - 1}}{ln ( 2 )}

= \frac{2 ^{t^{2} - 1}}{ln ( 2 )}

Add a constant to the solution

= \frac{2 ^{t^{2} - 1}}{ln ( 2 )} + C

\frac{d y}{d t} + y = e^{- t}

\frac{\partial}{\partial x} (\frac{1}{( 1 - x )})

\int \frac{x ^{2} + 9 x + 1}{( x ^{2} + 1 ) ^{2}} d x

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

\frac{\partial}{\partial x} (8 x^{3} - 24 x y + y^{3})

What is the integral of t2^{t^2} ?
The integral of t2^{t^2} is (2^{t^2-1})/(ln(2))+C