What is the integral of (40x)/(5x^2+e) ?

The integral of (40x)/(5x^2+e) is 4ln|5x^2+e|+C

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integral of (40x)/(5x^2+e)

\int \frac{40 x}{5 x ^{2} + e} d x

4 ln 5 x^{2} + e + C

Solution steps

\int \frac{40 x}{5 x ^{2} + e} d x

Take the constant out:

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

= 40 \cdot \int \frac{x}{5 x ^{2} + e} d x

Apply u-substitution

: \int \frac{1}{10 u} d u

= 40 \cdot \int \frac{1}{10 u} d u

Take the constant out:

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

= 40 \cdot \frac{1}{10} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 40 \cdot \frac{1}{10} ln ∣ u ∣

Substitute back

u = 5 x^{2} + e

= 40 \cdot \frac{1}{10} ln 5 x^{2} + e

Simplify

40 \cdot \frac{1}{10} ln 5 x^{2} + e : 4 ln 5 x^{2} + e

= 4 ln 5 x^{2} + e

Add a constant to the solution

= 4 ln 5 x^{2} + e + C

x \to 0 + lim (tan (x) ln (x))

\frac{d}{d x} (3 x + x)

y^{^{''}} + 4 y^{^{'}} + 5 y = 5 x + 3 e - x

x \to 1 lim (h (x))

\frac{\partial}{\partial x} (sin (yz))

What is the integral of (40x)/(5x^2+e) ?
The integral of (40x)/(5x^2+e) is 4ln|5x^2+e|+C