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

The integral of 1/((4x^2-25)^2) is

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

\int \frac{1}{( 4 x ^{2} - 25 ) ^{2}} d x

\frac{4 x ^{2} ln \frac{2 x}{5} + 1 - 4 x ^{2} ln \frac{2 x}{5} - 1 - 20 x + 25 ln \frac{2 x}{5} - 1 - 25 ln \frac{2 x}{5} + 1}{1000 ( 2 x + 5 ) ( 2 x - 5 )} + C

Solution steps

\int \frac{1}{( 4 x ^{2} - 25 ) ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{250 ( u ^{2} - 1 ) ^{2}} d u

Take the constant out:

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

= \frac{1}{250} \cdot \int \frac{1}{( u ^{2} - 1 ) ^{2}} d u

(u^{2} - 1)^{2} = u^{4} - 2 u^{2} + 1

= \frac{1}{250} \cdot \int \frac{1}{u ^{4} - 2 u ^{2} + 1} d u

Take the partial fraction of

\frac{1}{u ^{4} - 2 u ^{2} + 1} : \frac{1}{4 ( u + 1 )} + \frac{1}{4 ( u + 1 ) ^{2}} - \frac{1}{4 ( u - 1 )} + \frac{1}{4 ( u - 1 ) ^{2}}

= \frac{1}{250} \cdot \int \frac{1}{4 ( u + 1 )} + \frac{1}{4 ( u + 1 ) ^{2}} - \frac{1}{4 ( u - 1 )} + \frac{1}{4 ( u - 1 ) ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{250} (\int \frac{1}{4 ( u + 1 )} d u + \int \frac{1}{4 ( u + 1 ) ^{2}} d u - \int \frac{1}{4 ( u - 1 )} d u + \int \frac{1}{4 ( u - 1 ) ^{2}} d u)

\int \frac{1}{4 ( u + 1 )} d u = \frac{1}{4} ln ∣ u + 1 ∣

\int \frac{1}{4 ( u + 1 ) ^{2}} d u = - \frac{1}{4 ( u + 1 )}

\int \frac{1}{4 ( u - 1 )} d u = \frac{1}{4} ln ∣ u - 1 ∣

\int \frac{1}{4 ( u - 1 ) ^{2}} d u = - \frac{1}{4 ( u - 1 )}

= \frac{1}{250} (\frac{1}{4} ln ∣ u + 1 ∣ - \frac{1}{4 ( u + 1 )} - \frac{1}{4} ln ∣ u - 1 ∣ - \frac{1}{4 ( u - 1 )})

Substitute back

u = \frac{2}{5} x

= \frac{1}{250} (\frac{1}{4} ln \frac{2}{5} x + 1 - \frac{1}{4 ( \frac{2}{5} x + 1 )} - \frac{1}{4} ln \frac{2}{5} x - 1 - \frac{1}{4 ( \frac{2}{5} x - 1 )})

Simplify

\frac{1}{250} (\frac{1}{4} ln \frac{2}{5} x + 1 - \frac{1}{4 ( \frac{2}{5} x + 1 )} - \frac{1}{4} ln \frac{2}{5} x - 1 - \frac{1}{4 ( \frac{2}{5} x - 1 )}) : \frac{4 x ^{2} ln \frac{2 x}{5} + 1 - 4 x ^{2} ln \frac{2 x}{5} - 1 - 20 x + 25 ln \frac{2 x}{5} - 1 - 25 ln \frac{2 x}{5} + 1}{1000 ( 2 x + 5 ) ( 2 x - 5 )}

= \frac{4 x ^{2} ln \frac{2 x}{5} + 1 - 4 x ^{2} ln \frac{2 x}{5} - 1 - 20 x + 25 ln \frac{2 x}{5} - 1 - 25 ln \frac{2 x}{5} + 1}{1000 ( 2 x + 5 ) ( 2 x - 5 )}

Add a constant to the solution

= \frac{4 x ^{2} ln \frac{2 x}{5} + 1 - 4 x ^{2} ln \frac{2 x}{5} - 1 - 20 x + 25 ln \frac{2 x}{5} - 1 - 25 ln \frac{2 x}{5} + 1}{1000 ( 2 x + 5 ) ( 2 x - 5 )} + C

x - 1, 2 x + 6, [- 2, 4]

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

3 y^{^{''}} - 16 y^{^{'}} + 5 y = 0, y (0) = - 3, y^{^{'}} (0) = 4

\frac{1}{( x - 4 ) ^{2}}

f (x) = (3 x^{2} + 6 x)^{3}