What is the integral of (2xe^{x^2})/(e^{x^2)-2} ?

The integral of (2xe^{x^2})/(e^{x^2)-2} is ln|e^{x^2}-2|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2xe^{x^2})/(e^{x^2)-2}

\int \frac{2 x e ^{x^{2}}}{e ^{x^{2}} - 2} d x

ln e^{x^{2}} - 2 + C

Solution steps

\int \frac{2 x e ^{x^{2}}}{e ^{x^{2}} - 2} d x

Take the constant out:

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

= 2 \cdot \int \frac{x e ^{x^{2}}}{e ^{x^{2}} - 2} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = e^{x^{2}} - 2

= 2 \cdot \frac{1}{2} ln e^{x^{2}} - 2

Simplify

2 \cdot \frac{1}{2} ln e^{x^{2}} - 2 : ln e^{x^{2}} - 2

= ln e^{x^{2}} - 2

Add a constant to the solution

= ln e^{x^{2}} - 2 + C

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

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

\int x^{2} (x^{3} + 1)^{9} d x

\int e^{y + x} d y

\int sin^{4} (a) x d x

What is the integral of (2xe^{x^2})/(e^{x^2)-2} ?
The integral of (2xe^{x^2})/(e^{x^2)-2} is ln|e^{x^2}-2|+C