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\int \frac{x}{\left ( x-3 \right )\sqrt{x+1}}dx

  • Option 1)

    -2\sqrt{x+1}-\frac{3}{2}ln\left |\frac{ \sqrt{x+1}-2}{ \sqrt{x+1}+2} \right |+c

  • Option 2)

    2\sqrt{x+1}-\frac{3}{2}ln\left |\frac{ \sqrt{x+1}-2}{ \sqrt{x+1}+2} \right |+c

  • Option 3)

    2\sqrt{x+1}+\frac{3}{2}ln\left |\frac{ \sqrt{x+1}-2}{ \sqrt{x+1}+2} \right |+c

  • Option 4)

    none of these

 

Answers (1)

As we learnt

Type of Integration by perfect square -

The integration in the form 

(i) \int \frac{dx}{(px+q)\sqrt{ax^{2}+bx+c}}

(ii) \int \frac{dx}{(px+q)\sqrt{ax+b}}

(iii) \int \frac{(a+bx)^{m}}{(p+qx)^{n}}dx

- wherein

Working rule.

(i)  \rightarrow put  (px+q)=\frac{1}{t}

(ii) \rightarrow put  (ax+b)=t^{2}

(iii) \rightarrow put  (p+qx)=t

 

 Put x + 1 = t2. We get 

I = {\rm{ }}\int {\frac{{{\rm{(}}{{\rm{t}}^{\rm{2}}} - 1)2tdt}}{{({t^2} - 4)t}} = 2\int {\frac{{{t^2} - 1}}{{{t^2} - 4}}} dt}

= 2\int {\left\{ {1 + \frac{3}{{{t^2} - 4}}} \right\}dt{\rm{ }} = 2t + \frac{3}{2}\,\ln \left| {\frac{{t - 2}}{{t + 2}}} \right| + c}

- 2\sqrt {x + 1} - \frac{3}{2}\,\ln \left| {\frac{{\sqrt {x + 1} - 2}}{{\sqrt {x + 1} + 2}}} \right| + c


Option 1)

-2\sqrt{x+1}-\frac{3}{2}ln\left |\frac{ \sqrt{x+1}-2}{ \sqrt{x+1}+2} \right |+c

Option 2)

2\sqrt{x+1}-\frac{3}{2}ln\left |\frac{ \sqrt{x+1}-2}{ \sqrt{x+1}+2} \right |+c

Option 3)

2\sqrt{x+1}+\frac{3}{2}ln\left |\frac{ \sqrt{x+1}-2}{ \sqrt{x+1}+2} \right |+c

Option 4)

none of these

Posted by

subam

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