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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Evaluate:$\large \int\frac{x+2}{\sqrt{x^2+2x+3}}$dx

(Note: This question has been split into 2 questions) This question appeared in 65-1,65-2 and 65-3 versions of the paper in 2013.
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Toolbox:
  • When the given integral is of the form $\large\int\frac{px+q}{\sqrt{ax^2+bx+c}}$, we need to covert this using $px+q = A \large \frac{d(ax^2+bx+c)}{dx}$$+B$ and finding the values of $A$ and $B$.
  • When we see $ \large \int \frac{dx}{\sqrt(ax^2+bx+c)}$, we can rewrite this in the form $\large \int \frac{dx}{\sqrt(x+A)^2+B^2)}$
  • When we see $\large \int \frac{dx}{\sqrt (x^2+\alpha^2)}\; $we can apply the known result $=log \; |\;x+\sqrt(x^2+\alpha^2)\;| + k$
Given $I = \large \int\frac{x+2}{\sqrt{x^2+2x+3}}$$dx$
$\textbf{Step 1}$:
When the given integral is of the form $\large \int\frac{px+q}{\sqrt{ax^2+bx+c}}$, we need to covert this using $px+q = A \large \frac{d(ax^2+bx+c)}{dx}$$+B$ and finding the values of $A$ and $B$.
$\Rightarrow x+2 = A \large \frac{d(x^2+2x+3)}{dx}$$+B = 2Ax + 2A + B$
Equating the two terms, we get $2Ax = x$ and $2A+B = 2$.
$\Rightarrow 2Ax = x \rightarrow A = \large \frac{1}{2}$
$\Rightarrow B = 2 - 2A = 2 - 2\large\frac{1}{2}$$ = 2-1 = 1$
$\Rightarrow$ Therefore we can rewrite $x+2$ as $ A\large\frac{dt}{dx}$$+B$
$\textbf{Step 2}$:
Substituting for $t$ and $x+2$, we can rewrite the integral as $I = \large \int \frac {A\Large\frac{dt}{dx}}{\sqrt t}$$dx+\large \int \frac{B}{\sqrt t}$$dx$
$\Rightarrow I = A \large \int \frac{dt}{\sqrt t}$ $dx+ B\large \int \frac{1}{\sqrt t}$$dx$
$\Rightarrow I = \frac{1}{2} \large \int \frac{dt}{\sqrt t} + \int \frac{dx}{\sqrt(2x^2+2x+3)}$
Let us integrate them separetely, i.e.: $I_1 =\frac{1}{2} \large \int \frac{dt}{\sqrt t}$ and $I_2 = \large \int \frac{dx}{\sqrt(x^2+2x+3)}$
$\textbf{Step 3}$:
Given $I_1 =\frac{1}{2} \large \int \frac{dt}{\sqrt t}$
$\Rightarrow I_1 = \frac{1}{2} \large \int $$t^{\frac{-1}{2}} dt$ $= \frac{1}{2}\;2\; t^{\frac{1}{2}}$ $=\sqrt t$
$\Rightarrow I_1 = \sqrt (x^2+2x+3)$
$\textbf{Step 4}$:
$I_2 = \large \int \frac{dx}{\sqrt(x^2+2x+3)}$
When we see $ \large \int \frac{dx}{\sqrt(ax^2+bx+c)}$, we can rewrite this in the form $\large \int \frac{dx}{\sqrt(x+A)^2+B^2)}$
$ \Rightarrow$ Rewriting$\; x^2+2x+3 = x^2+2x+1+2 = (x+1)^2+(\sqrt2)^2$
$\Rightarrow$ $I_2 = \large \int \frac{dx}{\sqrt(x^2+2x+3)}$ $=\large \int \frac{dx}{\sqrt((x+1)^2+(\sqrt2)^2)}$
When we see $\large \int \frac{dx}{\sqrt (x^2+\alpha^2)}\; $we can apply the known result $=log \; |\;x+\sqrt(x^2+\alpha^2)\;| + k$
$\Rightarrow$ Therefore, $I_2 =\large \int \frac{dx}{\sqrt((x+1)^2+(\sqrt2)^2)}$ $= log|x+1+\sqrt((x+1)^2+\sqrt2)^2)| + k$
$\Rightarrow I_2 =log\;|\;x+1 + \sqrt (x^2+2x+3)\;| + k$
Therefore $\large \int\frac{x+2}{\sqrt{x^2+2x+3}}$$dx = \sqrt (x^2+2x+3) +log\;|\;x+1 + \sqrt (x^2+2x+3)\;| + k$
answered Mar 21, 2013 by sreemathi.v
edited Mar 22, 2013 by balaji.thirumalai
 

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