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

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

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

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

\frac{1}{2} arctan (2 t) + C

Solution steps

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

Apply Integral Substitution

= \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

= \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)

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

Substitute back

u = 2 t

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

Add a constant to the solution

= \frac{1}{2} arctan (2 t) + C

(x + y) d x + x^{2} d y = 0

\frac{\partial}{\partial x} (1 (x - y) \cdot e^{1 y + 4 x^{2}})

\frac{d ^{3}}{d x ^{3}} (6 x + 3)

\int_{0}^{s i n (x)} 8 t d t

\int - x - 3 d x

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