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Question: Solve \[\int{\dfrac{{{x}^{2}}-1}{{{x}^{3}}\sqrt{2{{x}^{4}}-2{{x}^{2}}+1}}dx}\] is equal to...

Solve ∫x2−1x32x4−2x2+1dx\int{\dfrac{{{x}^{2}}-1}{{{x}^{3}}\sqrt{2{{x}^{4}}-2{{x}^{2}}+1}}dx} is equal to

Explanation

Solution

This question belongs to the topic of integration which is from the integration. In solving this question, we will first solve the term x2−1x32x4−2x2+1\dfrac{{{x}^{2}}-1}{{{x}^{3}}\sqrt{2{{x}^{4}}-2{{x}^{2}}+1}} and make it simple so that the integration process will become easy. After that, we will use the substitution method of integration. After that, we will use of integration that is ∫xndx=xn+1n+1\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}. After solving the further process, we will get our answer.

Complete step-by-step solution:
Let us solve this question.
In this question, we have asked to do the integration. The term which we have to integrate with respect to x is x2−1x32x4−2x2+1\dfrac{{{x}^{2}}-1}{{{x}^{3}}\sqrt{2{{x}^{4}}-2{{x}^{2}}+1}}.
Now, let us first solve the term x2−1x32x4−2x2+1\dfrac{{{x}^{2}}-1}{{{x}^{3}}\sqrt{2{{x}^{4}}-2{{x}^{2}}+1}} and make it in simple form.
We can write x2−1x32x4−2x2+1\dfrac{{{x}^{2}}-1}{{{x}^{3}}\sqrt{2{{x}^{4}}-2{{x}^{2}}+1}} by dividing x5{{x}^{5}} to both numerator and denominator as
x2−1x32x4−2x2+1=x2−1x5x32x4−2x2+1x5\dfrac{{{x}^{2}}-1}{{{x}^{3}}\sqrt{2{{x}^{4}}-2{{x}^{2}}+1}}=\dfrac{\dfrac{{{x}^{2}}-1}{{{x}^{5}}}}{\dfrac{{{x}^{3}}\sqrt{2{{x}^{4}}-2{{x}^{2}}+1}}{{{x}^{5}}}}
The above can also be written as
⇒x2−1x32x4−2x2+1=x2x5−1x52x4−2x2+1x2\Rightarrow \dfrac{{{x}^{2}}-1}{{{x}^{3}}\sqrt{2{{x}^{4}}-2{{x}^{2}}+1}}=\dfrac{\dfrac{{{x}^{2}}}{{{x}^{5}}}-\dfrac{1}{{{x}^{5}}}}{\dfrac{\sqrt{2{{x}^{4}}-2{{x}^{2}}+1}}{{{x}^{2}}}}
The above equation can also be written as
⇒x2−1x32x4−2x2+1=1x3−1x52x4−2x2+1x2\Rightarrow \dfrac{{{x}^{2}}-1}{{{x}^{3}}\sqrt{2{{x}^{4}}-2{{x}^{2}}+1}}=\dfrac{\dfrac{1}{{{x}^{3}}}-\dfrac{1}{{{x}^{5}}}}{\dfrac{\sqrt{2{{x}^{4}}-2{{x}^{2}}+1}}{{{x}^{2}}}}
The above equation can also be written as
⇒x2−1x32x4−2x2+1=1x3−1x52x4−2x2+1x4\Rightarrow \dfrac{{{x}^{2}}-1}{{{x}^{3}}\sqrt{2{{x}^{4}}-2{{x}^{2}}+1}}=\dfrac{\dfrac{1}{{{x}^{3}}}-\dfrac{1}{{{x}^{5}}}}{\sqrt{\dfrac{2{{x}^{4}}-2{{x}^{2}}+1}{{{x}^{4}}}}}
The above equation can also be written as
⇒x2−1x32x4−2x2+1=1x3−1x52x4x4−2x2x4+1x4\Rightarrow \dfrac{{{x}^{2}}-1}{{{x}^{3}}\sqrt{2{{x}^{4}}-2{{x}^{2}}+1}}=\dfrac{\dfrac{1}{{{x}^{3}}}-\dfrac{1}{{{x}^{5}}}}{\sqrt{\dfrac{2{{x}^{4}}}{{{x}^{4}}}-\dfrac{2{{x}^{2}}}{{{x}^{4}}}+\dfrac{1}{{{x}^{4}}}}}
The above equation can also be written as
⇒x2−1x32x4−2x2+1=1x3−1x52−2x2+1x4\Rightarrow \dfrac{{{x}^{2}}-1}{{{x}^{3}}\sqrt{2{{x}^{4}}-2{{x}^{2}}+1}}=\dfrac{\dfrac{1}{{{x}^{3}}}-\dfrac{1}{{{x}^{5}}}}{\sqrt{2-\dfrac{2}{{{x}^{2}}}+\dfrac{1}{{{x}^{4}}}}}
Now, let's write the term which is inside the square root as t.
t=2−2x2+1x4t=2-\dfrac{2}{{{x}^{2}}}+\dfrac{1}{{{x}^{4}}}
Now, differentiating both sides, we get
dt=d(2)−d(2x2)+d(1x4)dt=d\left( 2 \right)-d\left( \dfrac{2}{{{x}^{2}}} \right)+d\left( \dfrac{1}{{{x}^{4}}} \right)
As we know that x to the power zero is 1, so we can write
⇒dt=2×d(x0)−2×d(x−2)+d(x−4)\Rightarrow dt=2\times d\left( {{x}^{0}} \right)-2\times d\left( {{x}^{-2}} \right)+d\left( {{x}^{-4}} \right)
Now, using the formula d(xn)=nxn−1dxd\left( {{x}^{n}} \right)=n{{x}^{n-1}}dx, we can write
⇒dt=0−2×(−2)×x−2−1d(x)+(−4)×x−4−1d(x)\Rightarrow dt=0-2\times \left( -2 \right)\times {{x}^{-2-1}}d\left( x \right)+\left( -4 \right)\times {{x}^{-4-1}}d\left( x \right)
The above can also be written as
⇒dt=4x−3dx−4x−5dx\Rightarrow dt=4{{x}^{-3}}dx-4{{x}^{-5}}dx
⇒dt4=x−3dx−x−5dx\Rightarrow \dfrac{dt}{4}={{x}^{-3}}dx-{{x}^{-5}}dx
⇒dt4=1x3dx−1x5dx\Rightarrow \dfrac{dt}{4}=\dfrac{1}{{{x}^{3}}}dx-\dfrac{1}{{{x}^{5}}}dx
⇒dt4=(1x3−1x5)dx\Rightarrow \dfrac{dt}{4}=\left( \dfrac{1}{{{x}^{3}}}-\dfrac{1}{{{x}^{5}}} \right)dx
Now, we can write the integration as
∫x2−1x32x4−2x2+1dx=∫(1x3−1x5)2−2x2+1x4dx=∫1tdt4=∫14tdt\int{\dfrac{{{x}^{2}}-1}{{{x}^{3}}\sqrt{2{{x}^{4}}-2{{x}^{2}}+1}}dx}=\int{\dfrac{\left( \dfrac{1}{{{x}^{3}}}-\dfrac{1}{{{x}^{5}}} \right)}{\sqrt{2-\dfrac{2}{{{x}^{2}}}+\dfrac{1}{{{x}^{4}}}}}dx}=\int{\dfrac{1}{\sqrt{t}}\dfrac{dt}{4}}=\int{\dfrac{1}{4\sqrt{t}}dt}
⇒∫x2−1x32x4−2x2+1dx=∫14tdt=14∫t−12dt\Rightarrow \int{\dfrac{{{x}^{2}}-1}{{{x}^{3}}\sqrt{2{{x}^{4}}-2{{x}^{2}}+1}}dx}=\int{\dfrac{1}{4\sqrt{t}}dt}=\dfrac{1}{4}\int{{{t}^{-\dfrac{1}{2}}}dt}
Now, using the formula ∫xndx=xn+1n+1\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}, we can write
⇒∫x2−1x32x4−2x2+1dx=14t−12+1−12+1\Rightarrow \int{\dfrac{{{x}^{2}}-1}{{{x}^{3}}\sqrt{2{{x}^{4}}-2{{x}^{2}}+1}}dx}=\dfrac{1}{4}\dfrac{{{t}^{-\dfrac{1}{2}+1}}}{-\dfrac{1}{2}+1}
The above can also be written as
⇒∫x2−1x32x4−2x2+1dx=14t1212=12t\Rightarrow \int{\dfrac{{{x}^{2}}-1}{{{x}^{3}}\sqrt{2{{x}^{4}}-2{{x}^{2}}+1}}dx}=\dfrac{1}{4}\dfrac{{{t}^{\dfrac{1}{2}}}}{\dfrac{1}{2}}=\dfrac{1}{2}\sqrt{t}
Now, substituting the value of t from above, we get
⇒∫x2−1x32x4−2x2+1dx=122−2x2+1x4\Rightarrow \int{\dfrac{{{x}^{2}}-1}{{{x}^{3}}\sqrt{2{{x}^{4}}-2{{x}^{2}}+1}}dx}=\dfrac{1}{2}\sqrt{2-\dfrac{2}{{{x}^{2}}}+\dfrac{1}{{{x}^{4}}}}
Hence, we have solved ∫x2−1x32x4−2x2+1dx\int{\dfrac{{{x}^{2}}-1}{{{x}^{3}}\sqrt{2{{x}^{4}}-2{{x}^{2}}+1}}dx} get the value as 122−2x2+1x4+C\dfrac{1}{2}\sqrt{2-\dfrac{2}{{{x}^{2}}}+\dfrac{1}{{{x}^{4}}}}+C, where C is constant.

Note: We should have a better knowledge in the topic of integration to solve this type of question easily. We should know that differentiation of constant is zero. We should remember the following formulas to solve this type of question easily:
∫xndx=xn+1n+1\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}
d(xn)=nxn−1dxd\left( {{x}^{n}} \right)=n{{x}^{n-1}}dx