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

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

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

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

4 arctan (t) + C

Solution steps

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

Use the common integral:

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

= 4 arctan (t)

Add a constant to the solution

= 4 arctan (t) + C

\int \frac{5}{t + 1} d t

\int \frac{1}{16 - 25 x ^{2}} d x

\frac{5}{x} - \frac{2}{x ^{3}} + \frac{1}{x ^{4}}

\int (4 x^{2} + 8 x + 9)^{6} (x + 1) d x

arccos (x)

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