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Question: How do you use substitution to integrate \(\dfrac{x}{{1 - x}}\)?...

How do you use substitution to integrate x1−x\dfrac{x}{{1 - x}}?

Explanation

Solution

In this question, we need to find the integration of the given function. Firstly, to make integration easier, we use substitution. Here we take t=1−xt = 1 - x and differentiate it with respect to x. Then using the expression of dxdx obtained, we integrate the given function. Now we substitute x=1−tx = 1 - t in the place of x. Then we integrate the obtained expression. We then substitute back t=1−xt = 1 - x and simplify to get the desired result.

Complete step by step solution:
Here we are asked to integrate the function x1−x\dfrac{x}{{1 - x}}.
We need to integrate the function using the substitution method.
So we find out ∫x1−xdx\int {\dfrac{x}{{1 - x}}dx} …… (1)
Firstly, we take 1−x1 - x some variable say t and proceed. i.e. take t=1−xt = 1 - x.
Now differentiating this with respect to x we get,
⇒dtdx=d(1−x)dx\Rightarrow \dfrac{{dt}}{{dx}} = \dfrac{{d(1 - x)}}{{dx}}
⇒dtdx=−dxdx\Rightarrow \dfrac{{dt}}{{dx}} = - \dfrac{{dx}}{{dx}}
⇒dtdx=−1\Rightarrow \dfrac{{dt}}{{dx}} = - 1
Now taking dxdx to the other side, we get,
⇒dt=−dx\Rightarrow dt = - dx
So the expression for dxdx is,
⇒dx=−dt\Rightarrow dx = - dt
Substituting t=1−xt = 1 - x in the equation (1), we get,
∫x1−xdx=∫xtdx\int {\dfrac{x}{{1 - x}}dx} = \int {\dfrac{x}{t}dx}
Now put dx=−dtdx = - dt, we get,
⇒∫x1−xdx=∫xt(−dt)\Rightarrow \int {\dfrac{x}{{1 - x}}} dx = \int {\dfrac{x}{t}} ( - dt)
We still have the variable x in the expression above. So let us use t substitution to solve for x.
We have taken t=1−xt = 1 - x
Hence the expression for x is, x=1−tx = 1 - t
Therefore, we get,
⇒∫x1−xdx=∫1−tt(−1)dt\Rightarrow \int {\dfrac{x}{{1 - x}}} dx = \int {\dfrac{{1 - t}}{t}} ( - 1)dt
Since −1 - 1 is a constant, from the constant coefficient rule we can take it out of integration.
Hence we have,
⇒∫x1−xdx=(−1)∫1−ttdt\Rightarrow \int {\dfrac{x}{{1 - x}}dx} = ( - 1)\int {\dfrac{{1 - t}}{t}dt}
⇒∫11−xdx=−∫(1t−tt)dt\Rightarrow \int {\dfrac{1}{{1 - x}}} dx = - \int {\left( {\dfrac{1}{t} - \dfrac{t}{t}} \right)} dt
Simplifying this, we get,
⇒∫x1−xdx=−∫(1t)dt+∫(tt)dt\Rightarrow \int {\dfrac{x}{{1 - x}}dx} = - \int {\left( {\dfrac{1}{t}} \right)} dt + \int {\left( {\dfrac{t}{t}} \right)} dt
⇒∫x1−xdx=−∫(1t)dt+∫1dt\Rightarrow \int {\dfrac{x}{{1 - x}}dx} = - \int {\left( {\dfrac{1}{t}} \right)} dt + \int {1dt}
We know that ∫1udu=ln⁡∣u∣+C\int {\dfrac{1}{u}} du = \ln |u| + C
Hence we have,
⇒∫x1−xdx=−(ln⁡∣t∣+t)+C\Rightarrow \int {\dfrac{x}{{1 - x}}dx} = - (\ln |t| + t) + C, where CC is an integration constant.
Substituting back t=1−xt = 1 - x we get,
⇒∫x1−xdx=−(ln⁡∣1−x∣+(1−x))+C\Rightarrow \int {\dfrac{x}{{1 - x}}dx} = - (\ln |1 - x| + (1 - x)) + C
⇒∫x1−xdx=−ln⁡∣1−x∣−1+x+C\Rightarrow \int {\dfrac{x}{{1 - x}}dx} = - \ln |1 - x| - 1 + x + C

Hence, the integration of x1−x\dfrac{x}{{1 - x}} is given by −ln⁡∣1−x∣−1+x+C - \ln |1 - x| - 1 + x + C, where CC is an integration constant.

Note: Students must remember that antiderivative is nothing but integration. And it is important to substitute 1−x1 - x as some variable, since it makes us to integrate easier and also it avoids confusion. Note that since in this problem the limits are not given. Hence the above problem is an example of indefinite integral. If the limits are given to apply, then we say that definite integral. Since the given problem is an indefinite integral, we have an integration constant after integrating the terms. This is important to add in such problems.
Also remember that ∫1udu=ln⁡∣u∣+C\int {\dfrac{1}{u}} du = \ln |u| + C