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Question

Question: $\int \frac{(1-t)^2}{(1+t)^4} dt$...

∫(1−t)2(1+t)4dt\int \frac{(1-t)^2}{(1+t)^4} dt

Answer

−16(1−t1+t)3+C-\frac{1}{6}\left(\frac{1-t}{1+t}\right)^3 + C

Explanation

Solution

Let the integral be II.

I=∫(1−t)2(1+t)4dtI = \int \frac{(1-t)^2}{(1+t)^4} dt

We can rewrite the integrand as: (1−t)2(1+t)4=(1−t)2(1+t)2(1+t)2=(1−t1+t)21(1+t)2\frac{(1-t)^2}{(1+t)^4} = \frac{(1-t)^2}{(1+t)^2 (1+t)^2} = \left(\frac{1-t}{1+t}\right)^2 \frac{1}{(1+t)^2}

Let's use the substitution v=1−t1+tv = \frac{1-t}{1+t}. To find dvdv, we differentiate vv with respect to tt using the quotient rule: dvdt=ddt(1−t)⋅(1+t)−(1−t)⋅ddt(1+t)(1+t)2\frac{dv}{dt} = \frac{\frac{d}{dt}(1-t) \cdot (1+t) - (1-t) \cdot \frac{d}{dt}(1+t)}{(1+t)^2}

dvdt=(−1)(1+t)−(1−t)(1)(1+t)2\frac{dv}{dt} = \frac{(-1)(1+t) - (1-t)(1)}{(1+t)^2}

dvdt=−1−t−(1−t)(1+t)2\frac{dv}{dt} = \frac{-1-t - (1-t)}{(1+t)^2}

dvdt=−1−t−1+t(1+t)2\frac{dv}{dt} = \frac{-1-t - 1+t}{(1+t)^2}

dvdt=−2(1+t)2\frac{dv}{dt} = \frac{-2}{(1+t)^2}

From this, we can express dt(1+t)2\frac{dt}{(1+t)^2} in terms of dvdv:

dt=(1+t)2−2dvdt = \frac{(1+t)^2}{-2} dv

dt(1+t)2=−12dv\frac{dt}{(1+t)^2} = -\frac{1}{2} dv

Substitute v=1−t1+tv = \frac{1-t}{1+t} and dt(1+t)2=−12dv\frac{dt}{(1+t)^2} = -\frac{1}{2} dv into the integral:

I=∫(1−t1+t)21(1+t)2dtI = \int \left(\frac{1-t}{1+t}\right)^2 \frac{1}{(1+t)^2} dt

I=∫v2(−12dv)I = \int v^2 \left(-\frac{1}{2} dv\right)

I=−12∫v2dvI = -\frac{1}{2} \int v^2 dv

Now, integrate with respect to vv using the power rule ∫vndv=vn+1n+1+C\int v^n dv = \frac{v^{n+1}}{n+1} + C:

I=−12(v2+12+1)+CI = -\frac{1}{2} \left(\frac{v^{2+1}}{2+1}\right) + C

I=−12(v33)+CI = -\frac{1}{2} \left(\frac{v^3}{3}\right) + C

I=−16v3+CI = -\frac{1}{6} v^3 + C

Finally, substitute back v=1−t1+tv = \frac{1-t}{1+t}:

I=−16(1−t1+t)3+CI = -\frac{1}{6} \left(\frac{1-t}{1+t}\right)^3 + C

I=−(1−t)36(1+t)3+CI = -\frac{(1-t)^3}{6(1+t)^3} + C

This is the indefinite integral of the given function.