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Question: Integrate the following: \( \left( a \right)I = \int {\dfrac{1}{{{x^3}\left( {1 - x} \right)}}dx} ...

Integrate the following:
(a)I=∫1x3(1−x)dx\left( a \right)I = \int {\dfrac{1}{{{x^3}\left( {1 - x} \right)}}dx}
(b)I=∫1x2(1+2x)dx\left( b \right)I = \int {\dfrac{1}{{{x^2}\left( {1 + 2x} \right)}}dx}

Explanation

Solution

Hint: To solve this integration question we have to make it easier by rearranging that means by manipulating the question we have to transform in the form so that formula of integration can be applied.

Complete step-by-step answer:
We have given
(a)I=∫1x3(1−x)dx\left( a \right)I = \int {\dfrac{1}{{{x^3}\left( {1 - x} \right)}}dx}
I=∫1x3(1−x)dxI = \int {\dfrac{1}{{{x^3}\left( {1 - x} \right)}}dx}
We can replace numerator 1 as 1-x+x
On replacing we get,
I=∫1−x+xx3(1−x)dxI = \int {\dfrac{{1 - x + x}}{{{x^3}\left( {1 - x} \right)}}dx}
Now we can write above equation as
I=∫1−xx3(1−x)dx+∫xx3(1−x)dxI = \int {\dfrac{{1 - x}}{{{x^3}\left( {1 - x} \right)}}dx} + \int {\dfrac{x}{{{x^3}\left( {1 - x} \right)}}dx}
Now on cancel out we get,
I=∫1x3dx+∫1x2(1−x)dxI = \int {\dfrac{1}{{{x^3}}}dx} + \int {\dfrac{1}{{{x^2}\left( {1 - x} \right)}}dx}
Now again we can write the equation as
I=∫1x3dx+∫1−x+xx2(1−x)dxI = \int {\dfrac{1}{{{x^3}}}dx} + \int {\dfrac{{1 - x + x}}{{{x^2}\left( {1 - x} \right)}}dx}
On rearranging we get,
I=∫1x3dx+∫1−xx2(1−x)dx+∫xx2(1−x)dxI = \int {\dfrac{1}{{{x^3}}}dx} + \int {\dfrac{{1 - x}}{{{x^2}\left( {1 - x} \right)}}dx} + \int {\dfrac{x}{{{x^2}\left( {1 - x} \right)}}} dx
On cancel out we get,
I=∫1x3dx+∫1x2dx+∫1x(1−x)dxI = \int {\dfrac{1}{{{x^3}}}dx} + \int {\dfrac{1}{{{x^2}}}dx} + \int {\dfrac{1}{{x\left( {1 - x} \right)}}} dx
Again we can write it as
I=∫1x3dx+∫1x2dx+∫1−x+xx(1−x)dxI = \int {\dfrac{1}{{{x^3}}}dx} + \int {\dfrac{1}{{{x^2}}}dx} + \int {\dfrac{{1 - x + x}}{{x\left( {1 - x} \right)}}} dx
On expanding we get,
I=∫1x3dx+∫1x2dx+∫1−xx(1−x)dx+∫xx(1−x)dxI = \int {\dfrac{1}{{{x^3}}}dx} + \int {\dfrac{1}{{{x^2}}}dx} + \int {\dfrac{{1 - x}}{{x\left( {1 - x} \right)}}dx + \int {\dfrac{x}{{x\left( {1 - x} \right)}}dx} }
On cancel out we get,
I=∫1x3dx+∫1x2dx+∫1xdx+∫1(1−x)dxI = \int {\dfrac{1}{{{x^3}}}dx} + \int {\dfrac{1}{{{x^2}}}dx} + \int {\dfrac{1}{x}dx + \int {\dfrac{1}{{\left( {1 - x} \right)}}dx} }
Now we will use the formula of integration to proceed further
(∫xndx=xn+1n+1)(∫1xdx=ln⁡x)(∫11−xdx=−ln⁡(1−x))\left( {\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} } \right)\left( {\int {\dfrac{1}{x}dx = \ln x} } \right)\left( {\int {\dfrac{1}{{1 - x}}dx = - \ln \left( {1 - x} \right)} } \right)
Using these formulae we get,
I=x−3+1−3+1+x−2+1−2+1+ln⁡x−ln⁡(1−x)+cI = \dfrac{{{x^{ - 3 + 1}}}}{{ - 3 + 1}} + \dfrac{{{x^{ - 2 + 1}}}}{{ - 2 + 1}} + \ln x - \ln \left( {1 - x} \right) + c
So final answer is
I=x−2−2−x−1+ln⁡x−ln⁡(1−x)+cI = \dfrac{{{x^{ - 2}}}}{{ - 2}} - {x^{ - 1}} + \ln x - \ln \left( {1 - x} \right) + c
Now we have second question:

(b)I=∫1x2(1+2x)dx\left( b \right)I = \int {\dfrac{1}{{{x^2}\left( {1 + 2x} \right)}}dx}
As we have solved first question same we will rearrange so that any formula can be applied
So we will rewrite the question as
I=∫1+2x−2xx2(1+2x)dxI = \int {\dfrac{{1 + 2x - 2x}}{{{x^2}\left( {1 + 2x} \right)}}dx}
On expanding we get,
I=∫1+2xx2(1+2x)dx−∫2xx2(1+2x)dxI = \int {\dfrac{{1 + 2x}}{{{x^2}\left( {1 + 2x} \right)}}dx} - \int {\dfrac{{2x}}{{{x^2}\left( {1 + 2x} \right)}}dx}
On cancel out we get,
I=∫1x2dx−∫2x(1+2x)dxI = \int {\dfrac{1}{{{x^2}}}dx} - \int {\dfrac{2}{{x\left( {1 + 2x} \right)}}dx}
Now we have to break or shorten ∫2x(1+2x)dx\int {\dfrac{2}{{x\left( {1 + 2x} \right)}}dx} this term so that it can be integrated easily.
We may write 1x(1+2x)=Ax+B(1+2x) 1=A(1+2x)+B(x) ∴2A+B=0,A=1 ∴B=−2  \dfrac{1}{{x\left( {1 + 2x} \right)}} = \dfrac{A}{x} + \dfrac{B}{{\left( {1 + 2x} \right)}} \\\ 1 = A\left( {1 + 2x} \right) + B\left( x \right) \\\ \therefore 2A + B = 0,A = 1 \\\ \therefore B = - 2 \\\
So now we have the required format which can be integrated easily.
We have ∫2x(1+2x)dx\int {\dfrac{2}{{x\left( {1 + 2x} \right)}}dx} = ∫2(1x−2(1+2x))dx\int {} 2\left( {\dfrac{1}{x} - \dfrac{2}{{\left( {1 + 2x} \right)}}} \right)dx
So finally our question becomes
I=∫1x2dx−I = \int {\dfrac{1}{{{x^2}}}dx} - ∫2(1x−2(1+2x))dx\int {} 2\left( {\dfrac{1}{x} - \dfrac{2}{{\left( {1 + 2x} \right)}}} \right)dx
Now using formulae (∫xndx=xn+1n+1)(∫1xdx=ln⁡x)(∫11−xdx=−ln⁡(1−x))\left( {\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} } \right)\left( {\int {\dfrac{1}{x}dx = \ln x} } \right)\left( {\int {\dfrac{1}{{1 - x}}dx = - \ln \left( {1 - x} \right)} } \right) we will get answer.
I=x−2+1−2+1−2ln⁡x+4ln⁡(1+2x)2 I=−1x−2ln⁡x+2ln⁡(1+2x)+c  I = \dfrac{{{x^{ - 2 + 1}}}}{{ - 2 + 1}} - 2\ln x + 4\dfrac{{\ln \left( {1 + 2x} \right)}}{2} \\\ I = \dfrac{{ - 1}}{x} - 2\ln x + 2\ln \left( {1 + 2x} \right) + c \\\

Note:- Whenever we get this type of question the key concept of solving is we have to have knowledge of integration formulae and also remember (∫xndx=xn+1n+1)(∫1xdx=ln⁡x)(∫11−xdx=−ln⁡(1−x))\left( {\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} } \right)\left( {\int {\dfrac{1}{x}dx = \ln x} } \right)\left( {\int {\dfrac{1}{{1 - x}}dx = - \ln \left( {1 - x} \right)} } \right) these so that we can solve integration questions easily.