What is the integral of (2u)/(u^2+1) ?

The integral of (2u)/(u^2+1) is ln|u^2+1|+C

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integral of (2u)/(u^2+1)

\int \frac{2 u}{u ^{2} + 1} d u

ln u^{2} + 1 + C

Solution steps

\int \frac{2 u}{u ^{2} + 1} d u

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 \cdot \frac{1}{2} ln ∣ v ∣

Substitute back

v = u^{2} + 1

= 2 \cdot \frac{1}{2} ln u^{2} + 1

Simplify

2 \cdot \frac{1}{2} ln u^{2} + 1 : ln u^{2} + 1

= ln u^{2} + 1

Add a constant to the solution

= ln u^{2} + 1 + C

166 y^{^{''}} + 135 y^{^{'}} + 1313 y = 36 \cdot 0.7 cos (x)

y^{^{''}} + a y + b = 0

\frac{d}{d x} (x y^{2} - y^{3})

n = 1 \sum \infty \frac{sin ( \frac{1}{n} )}{n}

n = 1 \sum \infty 4^{n + 1}

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