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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function\[\sqrt{x^2+3x}\]

$\begin{array}{1 1} \frac{(2x+3)(\sqrt{x^2+3x}) }{4}+\frac{9}{8} \log \mid \frac{2x+3}{2}+\sqrt{x^2+3x} \mid+c \\ \frac{(2x+3)(\sqrt{x^2+3x}) }{4}- \frac{7}{8} \log \mid \frac{2x+3}{4}+\sqrt{x^2+3x} \mid+c \\ \frac{(2x+3)(\sqrt{x^2+3x}) }{4}+\frac{7}{8} \log \mid \frac{2x+3}{4}+\sqrt{x^2-3x} \mid+c \\ \frac{(x+2)(\sqrt{x^2+3x}) }{2}-\frac{9}{8} \log \mid \frac{2x+3}{2}+\sqrt{x^2+3x} \mid+c \end{array} $

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  • ∫x2−a2dx=x2x2−a2+a22log|x+x2−a2|+c" role="presentation" style="position: relative;">x2a2dx=x2x2a2+a22log|x+x2a2|+c∫x2−a2dx=x2x2−a2+a22log|x+x2−a2|+c\int \sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}+\frac {a^2}{2}log |x+\sqrt{x^2-a^2}|+c
Given I=∫x2+3xdx" role="presentation" style="position: relative;">I=x2+3xdxI=∫x2+3xdxI=\int \sqrt{x^2+3x}dx
 
x2+3x" role="presentation" style="position: relative;">x2+3xx2+3x x^2+3x can written as
 
                  =x2+3x+94−94" role="presentation" style="position: relative;">=x2+3x+9494=x2+3x+94−94=x^2+3x+\frac{9}{4}-\frac{9}{4}
 
                  =(x+32)2−94" role="presentation" style="position: relative;">=(x+32)294=(x+32)2−94=(x+\frac{3}{2})^2-\frac{9}{4}
 
     Hence I=∫(x+32)2−(32)2dx" role="presentation" style="position: relative;">I=(x+32)2(32)2dxI=∫(x+32)2−(32)2dxI=\int \sqrt{(x+\frac{3}{2})^2-(\frac{3}{2})^2} dx
 
Clearly this is of the form ∫x2−a2dx" role="presentation" style="position: relative;">x2a2dx∫x2−a2dx\int \sqrt{x^2-a^2}dx
 
                  =x2x2−a2+a22log|x+x2−a2|dx" role="presentation" style="position: relative;">=x2x2a2+a22log|x+x2a2|dx=x2x2−a2+a22log|x+x2−a2|dx=\frac{x}{2}\sqrt{x^2-a^2}+\frac{a^2}{2}log|x+\sqrt{x^2-a^2}|dx
 
Here in the place of x we have (x+32)" role="presentation" style="position: relative;">(x+32)(x+32)(x+\frac{3}{2}) and in the place of a we have 32" role="presentation" style="position: relative;">3232\frac{3}{2}
 
Therefore ∫x2+3xdx=(x+32)2x2+3+94log|(x+32)+x2+3x|+c" role="presentation" style="position: relative;">x2+3xdx=(x+32)2x2+3+94log|(x+32)+x2+3x|+c∫x2+3xdx=(x+32)2x2+3+94log|(x+32)+x2+3x|+c\int \sqrt{x^2+3x} dx=\frac{(x+\frac{3}{2})}{2}\sqrt{x^2+3}+\frac{9}{4} log |(x+\frac{3}{2})+\sqrt{x^2+3x}|+c
 
                 =(2x+3)4x2+3x+98log|2x+32+x2+3x|+c" role="presentation" style="position: relative;">=(2x+3)4x2+3x+98log|2x+32+x2+3x|+c=(2x+3)4x2+3x+98log|2x+32+x2+3x|+c=\frac{(2x+3)}{4}\sqrt{x^2+3x}+\frac{9}{8}log|\frac{2x+3}{2}+\sqrt{x^2+3x}|+c
answered Feb 8, 2013 by meena.p
edited Apr 11 by meena.p
 
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