# How do you use partial fractions to find the integral #int (sec^2x)/(tanx(tanx+1))dx#?

Substitute:

so we have:

Now we can solve the integral using partial fractions:

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To integrate the given expression, we first rewrite it using partial fractions. We express ( \frac{\sec^2 x}{\tan x(\tan x + 1)} ) as the sum of two fractions:

[ \frac{\sec^2 x}{\tan x(\tan x + 1)} = \frac{A}{\tan x} + \frac{B}{\tan x + 1} ]

Next, we find the values of ( A ) and ( B ) by equating numerators:

[ \sec^2 x = A(\tan x + 1) + B(\tan x) ]

[ \sec^2 x = A\tan x + A + B\tan x ]

By rearranging terms, we have:

[ \sec^2 x = (A + B)\tan x + A ]

Comparing coefficients of ( \tan x ) and constants on both sides:

[ A + B = 0 ] [ A = 1 ]

From the first equation, we find ( B = -1 ).

So, we rewrite the integral as:

[ \int \frac{\sec^2 x}{\tan x(\tan x + 1)} dx = \int \frac{1}{\tan x} dx - \int \frac{1}{\tan x + 1} dx ]

Each of these integrals can be solved separately:

[ \int \frac{1}{\tan x} dx = \int \frac{\cos x}{\sin x} dx = \ln|\sin x| + C_1 ]

[ \int \frac{1}{\tan x + 1} dx = \int \frac{\cos x}{\sin x + \cos x} dx ]

To solve this, let ( u = \sin x + \cos x ), then ( du = (\cos x - \sin x) dx ), and rearrange for ( dx = \frac{du}{\cos x - \sin x} ):

[ \int \frac{1}{u} \frac{du}{\cos x - \sin x} ]

[ = \int \frac{1}{u} du ]

[ = \ln|u| + C_2 ]

[ = \ln|\sin x + \cos x| + C_2 ]

Therefore, the integral of ( \frac{\sec^2 x}{\tan x(\tan x + 1)} dx ) is:

[ \ln|\sin x| - \ln|\sin x + \cos x| + C ]

[ = \ln\left|\frac{\sin x}{\sin x + \cos x}\right| + C ]

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