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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Integrate : $\int \large\frac{1}{\sqrt {2x-3}- \sqrt {2x+1}}$$dx$

$(A)\;x^2+2x+3 \qquad(B)\;(2x+1)^{\frac{3}{2}} \qquad(C)\;(2x+3)^{\frac{3}{2}} \qquad (D)\;\frac{1}{6}(2x+1)^{\frac{3}{2}}+\frac{1}{6} (2x+1)^3+c$

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Multiply the numerator and denominator by $\sqrt {2x+3}+\sqrt {2x+1} $
=>$\int \large\frac{\sqrt {2x+3}+ \sqrt {2x+1}}{(\sqrt {2x-3}- \sqrt {2x+1})(\sqrt {2+3}+\sqrt {2x+1})}$$dx$
=>$\int \large\frac{\sqrt {2x+3}+ \sqrt {2x+1}}{2}$$dx$
=>$\large\frac{1}{2}$$ \int \sqrt {2x+3}dx+\large\frac{1}{2}$$ \int \sqrt {2x+1}$$dx$
=> $\large\frac{1}{2} \times \frac{1}{2}$$ (2x+3)^{3/2} \times \large\frac{2}{3} +\frac{1}{2} \times \frac{2}{3} \times \frac{1}{2}$$ (2x+1)^{\large\frac{3}{2}}$$+c$
=>$\large\frac{1}{6}$$(2x+1)^{\frac{3}{2}}+\frac{1}{6}$$ (2x+1)^3+c$
Hence d is the correct answer.
answered Dec 24, 2013 by meena.p
edited Mar 18, 2014 by balaji.thirumalai

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