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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Evaluate : $ \int \large\frac{2x+5}{\sqrt{7-6x-x^2}}$$dx$

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Toolbox:
  • $\int\large\frac{(px+q)}{\sqrt{ax^2+bx+c}}$$dx$,where p,q,a,b,c are constants,we are to find real numbers such that $px+q=A\frac{d}{dx}(ax^2+bx+c)+B.$
  • Therefore $px+q=A(2ax+b)+B.$
  • To determine A and B we equate the coefficients from both sided,the coefficients of x and the constant terms.A and B are thus obtained and hence the integral is reduced to one of the known forms
  • $\int \large\frac{dx}{\sqrt{a^2-x^2}}$$=-\sin^{-1}(\large\frac{x}{a})$
Step 1:
$I=\int\large\frac{2x+5}{\sqrt{7-6x-x^2}}$$dx$
Let us express
$2x+5=A\large\frac{d}{dx}$$(7-6x-x^2)+B$
$2x+5=A(-6-2x)+B$
Now equating the coefficients of $x$ and the constant terms from both sides we get,
$-2A=2$
$A=-1$
$-6A+B=5$
(i.e) $-6(-1)+B=5$
$6+B=5$
$B=-1$
Hence $A=-1$ and $B=-1$
$\therefore \int\large\frac{2x+5}{\sqrt{7-6x-x^2}}=\int\large\frac{-1(-6-2x)dx}{\sqrt{7-6x-x^2}}+\int\large\frac{-dx}{\sqrt{7-6x-x^2}}$
$\qquad\qquad\qquad=I_1-I_2$
Step 2:
In $I_1$ put $7-6x-x^2=t$
$\Rightarrow (-6-2x)dx=dt$
$\therefore I_1=-\int\large\frac{dt}{\sqrt t}$
$\qquad=-2\sqrt t+C_1$
$\qquad=-2\sqrt {7-6x-x^2}+C_1$
$I_2=-\int\large\frac{dx}{\sqrt{7-6x-x^2}}$
$\quad=-\int\large\frac{dx}{\sqrt{16-(x+3)^2}}$
$\int \large\frac{dx}{\sqrt{a^2-x^2}}$$=-\sin^{-1}(\large\frac{x}{a})$
$I_2=-\sin^{-1}(\large\frac{x+3}{4})$$+C_2$
Step 3:
$\therefore I=-(I_1+I_2)$
$\quad\;\;=-(2\sqrt{7-6x-x^2}+\sin^{-1}(\large\frac{x+3}{4}))$
answered Oct 2, 2013 by sreemathi.v
 
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