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

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

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

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

arctan (2 t) + C

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

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

Use the common integral:

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

= 2 \cdot \frac{1}{2} arctan (u)

Substitute back

u = 2 t

= 2 \cdot \frac{1}{2} arctan (2 t)

Simplify

2 \cdot \frac{1}{2} arctan (2 t) : arctan (2 t)

= arctan (2 t)

Add a constant to the solution

= arctan (2 t) + C

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