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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Find the integral $\int\sqrt x\;(3x^2+2x+3)\;dx$

$\begin{array}{1 1}\frac{6}{7}x^\frac{7}{2}+\frac{4}{5}x^\frac{5}{2}+2x^\frac{3}{2}+c. \\ \frac{6}{5}x^\frac{5}{2}-\frac{4}{3}x^\frac{3}{2}-2x^\frac{1}{2}+c. \\ \frac{6}{5}x^\frac{5}{2}+\frac{4}{3}x^\frac{3}{2}+2x^\frac{1}{2}+c. \\ \frac{6}{7}x^\frac{7}{2}-\frac{4}{5}x^\frac{5}{2}-2x^\frac{3}{2}+c. \end{array} $

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  • $(i)\int x^n dx=\frac{x^{n+1}}{n+1}+c$.
$\int\sqrt x(3x^2+2x+3)dx.$
 
On expanding we get,
 
$\int(3x^2\sqrt x+2x.\sqrt x+3\sqrt x)dx$.
 
$\;\;\;=3\int x^\frac{5}{2}dx+2\int x^\frac{3}{2}dx+3\int x^\frac{1}{2}dx.$
 
$\;\;\;=3\begin{bmatrix}\frac{x^\frac{5}{2}+1}{\frac{5}{2}+1}\end{bmatrix}+2\begin{bmatrix}\frac{x^\frac{3}{2}+1}{\frac{3}{2}+1}\end{bmatrix}+3\begin{bmatrix}\frac{x^\frac{1}{2}+1}{\frac{1}{2}+1}\end{bmatrix}$
 
$\;\;\;=3\begin{bmatrix}\frac{x^\frac{7}{2}}{\frac{7}{2}}\end{bmatrix}+2\begin{bmatrix}\frac{x^\frac{5}{2}}{\frac{5}{2}}\end{bmatrix}+3\begin{bmatrix}\frac{x^\frac{3}{2}}{\frac{3}{2}}\end{bmatrix}+c.$
 
$\;\;\;=\frac{6}{7}x^\frac{7}{2}+\frac{4}{5}x^\frac{5}{2}+2x^\frac{3}{2}+c.$
 
Hence $ \int\sqrt x(3x^2+2x+3)dx=\frac{6}{7}x^\frac{7}{2}+\frac{4}{5}x^\frac{5}{2}+2x^\frac{3}{2}+c.$

 

answered Jan 27, 2013 by sreemathi.v
 
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