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

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

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

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

ln t^{2} + 1 + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = t^{2} + 1

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

Simplify

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

= ln t^{2} + 1

Add a constant to the solution

= ln t^{2} + 1 + C

f (x) = ∣ 2 x - 3 ∣

f (x) = - sin (x^{2}) \cdot 2 x

\int \frac{x ^{4} - 3 x ^{2} + x - x + 2}{x} d x

7 + 3 t

\int \frac{x ^{3}}{( 1 - 2 x ^{4} ) ^{5}} d x

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