Mathematics

$$\int { \dfrac { 1 }{ \sqrt { 3x+5 } -\sqrt { 3x+2 }  }  } dx = $$


SOLUTION

We have,

$$\int{\dfrac{1}{\sqrt{3x+5}-\sqrt{3x+2}}dx}$$

On rationalize and we get,

  $$ \int{\dfrac{1}{\sqrt{3x+5}-\sqrt{3x+2}}\times \dfrac{\sqrt{3x+5}+\sqrt{3x+2}}{\sqrt{3x+5}+\sqrt{3x+2}}dx} $$

 $$ =\int{\dfrac{\sqrt{3x+5}+\sqrt{3x+2}}{{{\left( \sqrt{3x+5} \right)}^{2}}-{{\left( \sqrt{3x+2} \right)}^{2}}}}dx $$

 $$ =\int{\dfrac{\sqrt{3x+5}+\sqrt{3x+2}}{3x+5-3x-2}}dx $$

 $$ =\dfrac{1}{3}\int{\sqrt{3x+5}dx+\dfrac{1}{3}\int{\sqrt{3x+2}dx}} $$

 $$ =\dfrac{1}{3}{{\int{\left( 3x+5 \right)}}^{\dfrac{1}{2}}}dx+\dfrac{1}{3}{{\int{\left( 3x+2 \right)}}^{\dfrac{1}{2}}}dx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore \int{{{\left( ax+b \right)}^{n}}=}\dfrac{{{\left( ax+b \right)}^{n+1}}}{a\left( n+1 \right)} $$

 $$ =\dfrac{1}{3}\dfrac{{{\left( 3x+5 \right)}^{\dfrac{1}{2}+1}}}{3\left( \dfrac{1}{2}+1 \right)}+\dfrac{1}{3}\dfrac{{{\left( 3x+2 \right)}^{\dfrac{1}{2}+1}}}{3\left( \dfrac{1}{2}+1 \right)} $$

 $$ =\dfrac{1}{9}\dfrac{{{\left( 3x+5 \right)}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}+\dfrac{1}{9}\dfrac{{{\left( 3x+2 \right)}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} $$

 $$ =\dfrac{2}{9}\left[ {{\left( 3x+5 \right)}^{\dfrac{3}{2}}}+{{\left( 3x+2 \right)}^{\dfrac{3}{2}}} \right] $$

Hence, this is the answer.
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