What is the integral of (10x)/(5x^2+1) ?

The integral of (10x)/(5x^2+1) is ln|5x^2+1|+C

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integral of (10x)/(5x^2+1)

\int \frac{10 x}{5 x ^{2} + 1} d x

ln 5 x^{2} + 1 + C

Solution steps

\int \frac{10 x}{5 x ^{2} + 1} d x

Take the constant out:

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

= 10 \cdot \int \frac{x}{5 x ^{2} + 1} d x

Apply u-substitution

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

= 10 \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 10 \cdot \frac{1}{10} ln ∣ u ∣

Substitute back

u = 5 x^{2} + 1

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

Simplify

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

= ln 5 x^{2} + 1

Add a constant to the solution

= ln 5 x^{2} + 1 + C

\frac{d}{d x} (1 - \frac{1}{2 1 - x})

\frac{x ^{2} - 6 x + 1}{x ^{2} + 8}

5 + 7 x

ln (x e^{x} + 9)

f (x) = \frac{1}{x ^{2}}, at x = - 1

What is the integral of (10x)/(5x^2+1) ?
The integral of (10x)/(5x^2+1) is ln|5x^2+1|+C