What is the integral of (x^3+1)e^{x^4+4x} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^3+1)e^{x^4+4x}

\int (x^{3} + 1) e^{x^{4} + 4 x} d x

\frac{1}{4} e^{x^{4} + 4 x} + C

Solution steps

\int (x^{3} + 1) e^{x^{4} + 4 x} d x

Apply u-substitution

: \int \frac{e ^{u}}{4} d u

= \int \frac{e ^{u}}{4} d u

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{4} e^{u}

Substitute back

u = x^{4} + 4 x

= \frac{1}{4} e^{x^{4} + 4 x}

Add a constant to the solution

= \frac{1}{4} e^{x^{4} + 4 x} + C

\int \frac{8 x ^{4} - 9 x ^{3} + 6 x ^{2} - 2 x + 1}{x ^{2}} d x

\int (5 x^{2} + 2)^{6} 10 d x

\int cos^{3} (7 x) d x

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

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

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