What is the integral of (4t)/(1+t^2) ?

The integral of (4t)/(1+t^2) is 2ln|1+t^2|+C

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integral of (4t)/(1+t^2)

\int \frac{4 t}{1 + t ^{2}} d t

2 ln 1 + t^{2} + C

Solution steps

\int \frac{4 t}{1 + t ^{2}} d t

Take the constant out:

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

= 4 \cdot \int \frac{t}{1 + t ^{2}} d t

Apply u-substitution

= 4 \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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 + t^{2}

= 4 \cdot \frac{1}{2} ln 1 + t^{2}

Simplify

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

= 2 ln 1 + t^{2}

Add a constant to the solution

= 2 ln 1 + t^{2} + C

What is the integral of (4t)/(1+t^2) ?
The integral of (4t)/(1+t^2) is 2ln|1+t^2|+C