What is the integral of 1/(25x^2+16) ?

The integral of 1/(25x^2+16) is 1/20 arctan(5/4 x)+C

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integral of 1/(25x^2+16)

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

\frac{1}{20} arctan (\frac{5}{4} x) + C

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{20 ( 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}{20} \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}{20} arctan (u)

Substitute back

u = \frac{5}{4} x

= \frac{1}{20} arctan (\frac{5}{4} x)

Add a constant to the solution

= \frac{1}{20} arctan (\frac{5}{4} x) + C

x \to 2 lim (\frac{x}{x ^{2} - 3 x + 2})

\int \frac{- 9 x ^{2}}{4 x - x ^{2}} d x

\frac{d y}{d x} = e^{2 x} - 3 y

sin (\frac{π}{2})

y^{^{''}} + y = 8 e^{- 2 x}, y (0) = 0, y^{^{'}} (0) = 0

What is the integral of 1/(25x^2+16) ?
The integral of 1/(25x^2+16) is 1/20 arctan(5/4 x)+C