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\int \frac{dx}{\left ( x-b \right )^{3}\left ( x-a \right )^{2}}=

  • Option 1)

    -\frac{1}{\left ( a-b \right )^{4}}\left [ t-3ln\left | t \right |+\frac{3}{t} -\frac{1}{2t^{2}}\right ]\: \: where\: \: t=\frac{x-b}{x-a}

  • Option 2)

    -\frac{1}{\left ( a-b \right )^{4}}\left [ t-3ln\left | t \right |-\frac{3}{t} -\frac{1}{2t^{2}}\right ]\: \: where\: \: t=\frac{x-b}{x-a}

  • Option 3)

    -\frac{1}{\left ( a-b \right )^{4}}\left [ t-3ln\left | t \right |-\frac{3}{t} +\frac{1}{2t^{2}}\right ]\: \: where\: \: t=\frac{x-b}{x-a}

  • Option 4)

    -\frac{1}{\left ( a-b \right )^{4}}\left [ t+3ln\left | t \right |+\frac{3}{t} +\frac{1}{2t^{2}}\right ]\: \: where\: \: t=\frac{x-b}{x-a}

 

Answers (1)

As we learnt

Case for special type of indefinite integration -

\int x^{m}(a+bx^{n})^{p}dx

When P is an integer if P> 0 then apply expanded form

P< 0 then we put x=t^{k}

- wherein

Where k is the common denominator of m and n

 

 

Put x - b = t(x - a)

                   \therefore x=\frac{at-b}{t-1}\: \: \therefore dx=\frac{b-a}{\left ( t-1 \right )^{2}}dt

Also, x-a=\frac{a-b}{t-1}\: \: and\: \: x-b=\frac{\left ( a-b \right )t}{t-1}

Hence I=\int \frac{\left ( t-1 \right )^{3}}{\left ( a-b \right )^{3}t^{3}}.\frac{\left ( t-1 \right )^{2}}{\left ( a-b \right )^{2}}.\frac{b-a}{\left ( t-1 \right )^{2}}dt

   =-\frac{1}{\left ( a-b \right )^{4}}\int \frac{t^{3}-3t^{2}+3t-1}{t^{3}}dt=-\frac{1}{\left ( a-b \right )^{4}}\int \left ( 1-\frac{3}{t}+\frac{3}{t^{2}}-\frac{1}{t^{3}} \right )dt.=-\frac{1}{\left ( a-b \right )^{4}}\left [ t-3\ln \left | t \right |-\frac{3}{t} +\frac{1}{2t^{2}}\right ]\: where\: t=\frac{x-b}{x-a}


Option 1)

-\frac{1}{\left ( a-b \right )^{4}}\left [ t-3ln\left | t \right |+\frac{3}{t} -\frac{1}{2t^{2}}\right ]\: \: where\: \: t=\frac{x-b}{x-a}

Option 2)

-\frac{1}{\left ( a-b \right )^{4}}\left [ t-3ln\left | t \right |-\frac{3}{t} -\frac{1}{2t^{2}}\right ]\: \: where\: \: t=\frac{x-b}{x-a}

Option 3)

-\frac{1}{\left ( a-b \right )^{4}}\left [ t-3ln\left | t \right |-\frac{3}{t} +\frac{1}{2t^{2}}\right ]\: \: where\: \: t=\frac{x-b}{x-a}

Option 4)

-\frac{1}{\left ( a-b \right )^{4}}\left [ t+3ln\left | t \right |+\frac{3}{t} +\frac{1}{2t^{2}}\right ]\: \: where\: \: t=\frac{x-b}{x-a}

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