What is the integral of (xe^{x^2})/(1+e^{x^2)} ?

The integral of (xe^{x^2})/(1+e^{x^2)} is 1/2 ln|1+e^{x^2}|+C

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integral of (xe^{x^2})/(1+e^{x^2)}

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

\frac{1}{2} ln 1 + e^{x^{2}} + C

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 + e^{x^{2}}

= \frac{1}{2} ln 1 + e^{x^{2}}

Add a constant to the solution

= \frac{1}{2} ln 1 + e^{x^{2}} + C

f (x) = 2 x^{2} - 4 x

f (x) x^{2} + 8 x - 2

\int_{1}^{10} \frac{1}{x} d x

- arctan (x)

\int \frac{x e ^{5 x}}{( 5 x + 1 ) ^{2}} d x

What is the integral of (xe^{x^2})/(1+e^{x^2)} ?
The integral of (xe^{x^2})/(1+e^{x^2)} is 1/2 ln|1+e^{x^2}|+C