Question

The value of the integral $\int\limits_{0}^{\dfrac{1}{2}}{\dfrac{1+\sqrt{3}}{{{\left( {{\left( x+1 \right)}^{2}}{{\left( 1-x \right)}^{6}} \right)}^{\dfrac{1}{4}}}}dx}$

Answer 1

To solve this question, we should know how to arrange the terms in the given question, to get an easier form of integration. In the question, let us consider the denominator ${{\left( {{\left( x+1 \right)}^{2}}{{\left( 1-x \right)}^{6}} \right)}^{\dfrac{1}{4}}}$. By multiplying and dividing by ${{\left( 1-x \right)}^{2}}$ inside the power, we can change the denominator as ${{\left( \dfrac{{{\left( x+1 \right)}^{2}}{{\left( 1-x \right)}^{6}}{{\left( 1-x \right)}^{2}}}{{{\left( 1-x \right)}^{2}}} \right)}^{\dfrac{1}{4}}}$ and this can be further modified as ${{\left( \dfrac{{{\left( x+1 \right)}^{2}}}{{{\left( 1-x \right)}^{2}}} \right)}^{\dfrac{1}{4}}}{{\left( {{\left( 1-x \right)}^{8}} \right)}^{\dfrac{1}{4}}}$. The integral becomes $\int\limits_{0}^{\dfrac{1}{2}}{\dfrac{1+\sqrt{3}}{{{\left( \dfrac{x+1}{1-x} \right)}^{\dfrac{1}{2}}}{{\left( 1-x \right)}^{2}}}dx}$. Let us consider $\dfrac{x+1}{1-x}=t$ and by differentiating on both sides, we get
$\begin{aligned} & \dfrac{\left( 1-x \right)1-\left( x+1 \right)\left( -1 \right)}{{{\left( 1-x \right)}^{2}}}dx=dt \\\ & \dfrac{2}{{{\left( 1-x \right)}^{2}}}dx=dt \\\ \end{aligned}$
Using this substitution simplifies the integration and we can find the value of the integral.

Complete step-by-step answer:

We are given the integral $\int\limits_{0}^{\dfrac{1}{2}}{\dfrac{1+\sqrt{3}}{{{\left( {{\left( x+1 \right)}^{2}}{{\left( 1-x \right)}^{6}} \right)}^{\dfrac{1}{4}}}}dx}$.
We can see that the direct integral of the expression inside the integral is not easy and not possible. To do these types of integrals, we should either substitute a function such as $\cos \theta$ or $\tan \theta$ or modify the given expression and then substitute a whole function in x as another variable u. Mathematically, either substitute $x=f\left( \theta \right)$ or simplify the above integral by substituting $f\left( x \right)=u$.
In our case, we can see that we cannot find a trigonometric function in terms of $\theta$ which can simplify the given function. SO, we should use an alternative way of substitution to solve this integral.
Let us consider the denominator of the function which is ${{\left( {{\left( x+1 \right)}^{2}}{{\left( 1-x \right)}^{6}} \right)}^{\dfrac{1}{4}}}$.
Let us multiply and divide by ${{\left( 1-x \right)}^{2}}$ inside the power, we get
${{\left( \dfrac{{{\left( x+1 \right)}^{2}}{{\left( 1-x \right)}^{6}}{{\left( 1-x \right)}^{2}}}{{{\left( 1-x \right)}^{2}}} \right)}^{\dfrac{1}{4}}}$.
We can modify the above expression as ${{\left( \dfrac{{{\left( x+1 \right)}^{2}}}{{{\left( 1-x \right)}^{2}}} \right)}^{\dfrac{1}{4}}}{{\left( {{\left( 1-x \right)}^{8}} \right)}^{\dfrac{1}{4}}}={{\left( \dfrac{x+1}{1-x} \right)}^{\dfrac{1}{2}}}{{\left( 1-x \right)}^{2}}$
The integral becomes
$\int\limits_{0}^{\dfrac{1}{2}}{\dfrac{1+\sqrt{3}}{{{\left( \dfrac{x+1}{1-x} \right)}^{\dfrac{1}{2}}}{{\left( 1-x \right)}^{2}}}dx}$
To solve this integral, we should substitute the expression $\dfrac{x+1}{1-x}=t\to \left( 1 \right)$.
By differentiating the above expression, we get
$\begin{aligned} & \dfrac{\left( 1-x \right)1-\left( x+1 \right)\left( -1 \right)}{{{\left( 1-x \right)}^{2}}}dx=dt \\\ & \dfrac{2}{{{\left( 1-x \right)}^{2}}}dx=dt \\\ & \dfrac{1}{{{\left( 1-x \right)}^{2}}}dx=\dfrac{1}{2}dt\to \left( 2 \right) \\\ \end{aligned}$
Let us take the value of the lower limit x = 0
From equation-1 we get the lower limit of t as $\dfrac{0+1}{1-0}=t=1$
Let us take the value of the lower limit x = $\dfrac{1}{2}$
From equation-1 we get the lower limit of t as $\dfrac{\dfrac{1}{2}+1}{1-\dfrac{1}{2}}=\dfrac{\dfrac{3}{2}}{\dfrac{1}{2}}=3=t$

the equations-1 and 2 and the limits, we get
$\int\limits_{1}^{3}{\dfrac{1+\sqrt{3}}{{{t}^{\dfrac{1}{2}}}\times 2}dt}=\dfrac{1+\sqrt{3}}{2}\int\limits_{1}^{3}{\dfrac{1}{{{t}^{\dfrac{1}{2}}}}dt}=\dfrac{1+\sqrt{3}}{2}\int\limits_{1}^{3}{{{t}^{\dfrac{-1}{2}}}dt}$
We know that $\int{{{t}^{n}}dt=\dfrac{{{t}^{n+1}}}{n+1}}$
Using this formula, we get

& \left( \dfrac{1+\sqrt{3}}{2} \right)\int\limits_{1}^{3}{{{t}^{\dfrac{-1}{2}}}dt}=\left( \dfrac{1+\sqrt{3}}{2} \right)\left[ \dfrac{{{t}^{\dfrac{-1}{2}+1}}}{\dfrac{-1}{2}+1} \right]_{1}^{3}=\left( \dfrac{1+\sqrt{3}}{2} \right)\left[ \dfrac{{{t}^{\dfrac{1}{2}}}}{\dfrac{1}{2}} \right]_{1}^{3} \\\ & =\left( 1+\sqrt{3} \right)\left[ \sqrt{3}-\sqrt{1} \right]=\left( \sqrt{3}-1 \right)\left( \sqrt{3}+1 \right)=3-1=2 \\\ \end{aligned}$$ $\therefore $ The value of the integral $\int\limits_{0}^{\dfrac{1}{2}}{\dfrac{1+\sqrt{3}}{{{\left( {{\left( x+1 \right)}^{2}}{{\left( 1-x \right)}^{6}} \right)}^{\dfrac{1}{4}}}}dx}$ is 2. **Note:** Students might make a mistake while substituting the values of t and dt back in the equation. The most common mistake is by forgetting the value of 2 in $\dfrac{1}{{{\left( 1-x \right)}^{2}}}dx=\dfrac{1}{2}dt$. This leads to a value of the integral as 4 instead of 2. To avoid these mistakes, it is a good practice to modify the dx part of the integral into the form it looks in the relation between dx and dt. Another probable mistake is that students don’t change the limit and substitute the values of $0$ and $\dfrac{1}{2}$ in place of t. To avoid this kind of mistake, either substitute the value of t in terms of x back in the final result after integration and substitute limits in x or convert the limits into t as we did in this question.