What is the integral of 1/(25x^2+4) ?

The integral of 1/(25x^2+4) is 1/10 arctan(5/2 x)+C

Popular >

integral of 1/(25x^2+4)

\int \frac{1}{25 x ^{2} + 4} d x

\frac{1}{10} arctan (\frac{5}{2} x) + C

Solution steps

\int \frac{1}{25 x ^{2} + 4} d x

Apply Integral Substitution

= \int \frac{1}{10 ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{10} arctan (u)

Substitute back

u = \frac{5}{2} x

= \frac{1}{10} arctan (\frac{5}{2} x)

Add a constant to the solution

= \frac{1}{10} arctan (\frac{5}{2} x) + C

\int \frac{x x - 1}{1 - x - 1} d x

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

\int - 2 x^{4} + 3 x^{2} - 1 d x

\int \frac{1}{( x ^{2} - x + 1 ) ^{2}} d x

\int \frac{sin ( t )}{1 + sin ( t )} d t

What is the integral of 1/(25x^2+4) ?
The integral of 1/(25x^2+4) is 1/10 arctan(5/2 x)+C