What is the integral of x^33^{x^4} ?

The integral of x^33^{x^4} is 1/(4ln(3))*3^{x^4}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^33^{x^4}

\int x^{3} 3^{x^{4}} d x

\frac{1}{4 ln ( 3 )} \cdot 3^{x^{4}} + C

Solution steps

\int x^{3} \cdot 3^{x^{4}} d x

Apply u-substitution

: \int \frac{1}{4 ln ( 3 )} d u

= \int \frac{1}{4 ln ( 3 )} d u

Integral of a constant:

\int a d x = a x

= \frac{1}{4 ln ( 3 )} u

Substitute back

u = 3^{x^{4}}

= \frac{1}{4 ln ( 3 )} \cdot 3^{x^{4}}

Add a constant to the solution

= \frac{1}{4 ln ( 3 )} \cdot 3^{x^{4}} + C

\int - 5 y d y

\int cos (4 x) x d x

\int \frac{x ^{3}}{( 3 - 2 x ) ^{5}} d x

\int \frac{4}{( x ^{3} + 4 x )} d x

\int \frac{2 x}{9 + x ^{2}} d x

What is the integral of x^33^{x^4} ?
The integral of x^33^{x^4} is 1/(4ln(3))*3^{x^4}+C