What is the integral of (5xe^{2x^2})/(7e^{2x^2)+4} ?

The integral of (5xe^{2x^2})/(7e^{2x^2)+4} is 5/28 ln|7e^{2x^2}+4|+C

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integral of (5xe^{2x^2})/(7e^{2x^2)+4}

\int \frac{5 x e ^{2 x^{2}}}{7 e ^{2 x^{2}} + 4} d x

\frac{5}{28} ln 7 e^{2 x^{2}} + 4 + C

Solution steps

\int \frac{5 x e ^{2 x^{2}}}{7 e ^{2 x^{2}} + 4} d x

Take the constant out:

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

= 5 \cdot \int \frac{x e ^{2 x^{2}}}{7 e ^{2 x^{2}} + 4} d x

Apply u-substitution

: \int \frac{1}{28 u} d u

= 5 \cdot \int \frac{1}{28 u} d u

Take the constant out:

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

= 5 \cdot \frac{1}{28} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 \cdot \frac{1}{28} ln ∣ u ∣

Substitute back

u = 7 e^{2 x^{2}} + 4

= 5 \cdot \frac{1}{28} ln 7 e^{2 x^{2}} + 4

Simplify

5 \cdot \frac{1}{28} ln 7 e^{2 x^{2}} + 4 : \frac{5}{28} ln 7 e^{2 x^{2}} + 4

= \frac{5}{28} ln 7 e^{2 x^{2}} + 4

Add a constant to the solution

= \frac{5}{28} ln 7 e^{2 x^{2}} + 4 + C

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What is the integral of (5xe^{2x^2})/(7e^{2x^2)+4} ?
The integral of (5xe^{2x^2})/(7e^{2x^2)+4} is 5/28 ln|7e^{2x^2}+4|+C