What is the integral of 50t^9e^{5t^5} ?

The integral of 50t^9e^{5t^5} is 2e^{5t^5}t^5-2/5 e^{5t^5}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 50t^9e^{5t^5}

\int 50 t^{9} e^{5 t^{5}} d t

2 e^{5 t^{5}} t^{5} - \frac{2}{5} e^{5 t^{5}} + C

Solution steps

\int 50 t^{9} e^{5 t^{5}} d t

Take the constant out:

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

= 50 \cdot \int t^{9} e^{5 t^{5}} d t

Apply u-substitution

: \int \frac{e ^{u} u}{125} d u

= 50 \cdot \int \frac{e ^{u} u}{125} d u

Take the constant out:

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

= 50 \cdot \frac{1}{125} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 50 \cdot \frac{1}{125} (e^{u} u - \int e^{u} d u)

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

= 50 \cdot \frac{1}{125} (e^{u} u - e^{u})

Substitute back

u = 5 t^{5}

= 50 \cdot \frac{1}{125} (e^{5 t^{5}} \cdot 5 t^{5} - e^{5 t^{5}})

Simplify

50 \cdot \frac{1}{125} (e^{5 t^{5}} \cdot 5 t^{5} - e^{5 t^{5}}) : 2 e^{5 t^{5}} t^{5} - \frac{2}{5} e^{5 t^{5}}

= 2 e^{5 t^{5}} t^{5} - \frac{2}{5} e^{5 t^{5}}

Add a constant to the solution

= 2 e^{5 t^{5}} t^{5} - \frac{2}{5} e^{5 t^{5}} + C

\int (cot (x) + 3)^{2} d x

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

\int \frac{( 1 + ln ( x ) )}{x} d x

\int \frac{x + 1}{x ^{2} + 4 x + 6} d x

\int \frac{x tan ( x )}{cos ^{2} ( x )} d x

What is the integral of 50t^9e^{5t^5} ?
The integral of 50t^9e^{5t^5} is 2e^{5t^5}t^5-2/5 e^{5t^5}+C