Question

Integrate the given expression,
\[\int{\dfrac{{{x}^{2}}}{{{\left( a+bx \right)}^{2}}}dx,x\ne \dfrac{-a}{b}}\]

Answer 1

Hint: Put, \[a+bx=t\]. Thus from this formulate expression for x and put \[dx=dt\]. Substitute, simplify and integrate the expression formed. Then replace it by \[\left( a+bx \right)\].

Complete step-by-step answer:

We have been given an expression to integrate. Let us take it as I.

\[\therefore I=\int{\dfrac{{{x}^{2}}}{{{\left( a+bx \right)}^{2}}}dx}-(1)\]

Now let us put, \[a+bx=t\Rightarrow x=\dfrac{t-a}{b}\]

Hence, \[dx=dt\] and \[x=\left( \dfrac{t-a}{b} \right)\].

Now let us put these values in (1). Thus I becomes,

\[\begin{align}

  & I=\int{\dfrac{{{\left( \dfrac{t-a}{b} \right)}^{2}}}{{{t}^{2}}}.dt=\int{\dfrac{{{\left( t-a \right)}^{2}}}{{{b}^{2}}.{{t}^{2}}}.dt=\dfrac{1}{{{b}^{2}}}\int{\dfrac{{{\left( t-a \right)}^{2}}}{{{t}^{2}}}.dt}}} \\

 & \therefore I=\dfrac{1}{{{b}^{2}}}\int{\dfrac{{{\left( t-a \right)}^{2}}}{{{t}^{2}}}dt} \\

\end{align}\]

Now let us expand the numerator, \[{{\left( t-a \right)}^{2}}\] which I of the form, \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\].

\[\begin{align}

  & I=\dfrac{1}{{{b}^{2}}}\int{\dfrac{{{t}^{2}}-2at+{{a}^{2}}}{{{t}^{2}}}.dt} \\

 & I=\dfrac{1}{{{b}^{2}}}\int{\left( \dfrac{{{t}^{2}}}{{{t}^{2}}}-\dfrac{2at}{{{t}^{2}}}+\dfrac{{{a}^{2}}}{{{t}^{2}}} \right)dt} \\

\end{align}\]

Now let us split each term with the integral and integrate them separately.

\[\begin{align}

  & I=\dfrac{1}{{{b}^{2}}}\left[ \int{\dfrac{{{t}^{2}}}{{{t}^{2}}}dt}-\int{\dfrac{2at}{{{t}^{2}}}dt}+\int{\dfrac{{{a}^{2}}}{{{t}^{2}}}.

dt} \right] \\

 & I=\dfrac{1}{{{b}^{2}}}\left[ \int{1.dt}-a\int{\dfrac{2}{t}dt}+{{a}^{2}}\int{\dfrac{1}{{{t}^{2}}}dt} \right]-(2) \\

\end{align}\]

We know the basic integral formula,

\[\begin{align}

  & \int{1.dx}=x \\

 & \int{\dfrac{1}{x}.dx=\log x} \\

 & \int{\dfrac{1}{{{x}^{2}}}dx}=\int{{{x}^{-2}}.dx}=\dfrac{{{x}^{-2+1}}}{-2+1}=\dfrac{{{x}^{-1}}}{-1}

=\dfrac{-1}{x} \\

\end{align}\]

Now let us apply these basic formulas in equation (2) and simplify it.

\[\begin{align}

  & I=\dfrac{1}{{{b}^{2}}}\left[ t-2a\log t+{{a}^{2}}\left( \dfrac{{{t}^{-2+1}}}{-2+1} \right) \right]+C \\

 & I=\dfrac{1}{{{b}^{2}}}\left[ t-2a\log t+{{a}^{2}}\left( -{{t}^{-1}} \right) \right]+C \\

 & I=\dfrac{1}{{{b}^{2}}}\left[ t-2a\log t-\dfrac{{{a}^{2}}}{t} \right]+C-(3) \\

\end{align}\]

Now let us put, \[\left( a+bx \right)\] in place of t in equation (3).

\[I=\dfrac{1}{{{b}^{2}}}\left[ \left( a+bx \right)-2a\log \left( a+bx \right)-\dfrac{{{a}^{2}}}{\left( a+bx \right)} \right]+C\]

Thus the above expression is the required interval.

\[\int{\dfrac{{{x}^{2}}}{{{\left( a+bx \right)}^{2}}}dx}=\dfrac{1}{{{b}^{2}}}\left[ \left( a+bx \right)-2a\log \left( a+bx \right)-\dfrac{{{a}^{2}}}{\left( a+bx \right)} \right]+C\]

Hence, we got the required answer.


Note: It is important that you put, \[a+bx=t\], don’t elaborate or try to simplify it directly because then the solution becomes complex. Remember the basic integration identities that we have used here.