What is the integral of (k*x)/((1+x^2)) ?

The integral of (k*x)/((1+x^2)) is k 1/2 ln|1+x^2|+C

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integral of (k*x)/((1+x^2))

integral

\frac{k \cdot x}{( 1 + x ^{2} )}

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= k \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= k \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 + x^{2}

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

Add a constant to the solution

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

lo g_{10} (x + 2)

lo g_{10} (x + 1)

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

\frac{x}{( 1 + x ^{2} )}

\frac{x ^{2}}{( x + 1 ) ( 2 x + 3 )}

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