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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the following $\int\frac{3x-1}{\sqrt{x^2+9}}dx$

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Toolbox:
  • If a rational expression is of the form $\large\frac{px+q}{\sqrt {ax^2+bx+c}}$ then it can be resolved as $A\large\frac{d}{dx}$$(ax^2+bx+c)+B$
  • If $f(x)$ is substituted by $f(t),$ then $f;(x)dx=f'(t)dt$, hence $\int f(x)dx=\int f'(t)dt$
Step 1:
Let $I=\int \large\frac{3x-1}{\sqrt {x^2-9}}dx$
Let $3x-1=A \large\frac{d}{dx} $$(x^2-9)+B$
$(3x-1)=A(2x)+B$
To evaluate A and B, put x=0
$-1=B$
Next put $x=1$
$3(1)-1=A( 2 \times 1)+B$
$2=2A+B$
But $B=-1$
Therefore $2=2A-1$
$2A=3 =>A=\large\frac{3}{2}$
Therefore $A=\large\frac{3}{2}$ $;and \;B=-1$
Hence $\int \large\frac{3x-1}{\sqrt {x^2-9}}dx=\int \large\frac{\frac{3}{2}(2x)}{\sqrt {x^2-9}}dx+ \int \frac{(-1)dx}{\sqrt {x^2-9}}$
$= 3 \int \large\frac{xdx}{\sqrt {x^2-9}}-\int\frac{dx}{\sqrt {x^2-9}}$
Step 2:
consider $I_1=3 \int \large\frac{xdx}{\sqrt {x^2-9}}$
Put $x^2-9=t$
differentiating with respect to x we get
$2xdx=dt$
$xdx=\large\frac{dt}{2}$
substituting for $x^2-9$ and dx we get
$I_1=3 \int \large\frac{dt/2}{\sqrt t}=\frac{3}{2} \int \frac{dt}{\sqrt t}$
On integrating we get
$I_1=3 \bigg[\large\frac{1}{2}\bigg(\frac{t^{-1/2+1}}{-1/2+1}\bigg)\bigg]=+ \sqrt t$
Substituting for t we get
$I_1=3 \sqrt {x^2-9}$
Step 3:
Next let us consider $I_2=\int \large \frac{dx}{\sqrt {x^2-9}}$
This is of the form $\int\large\frac{dx}{\sqrt{x^2-a^2}}$$=\log|(x-a)+\sqrt {x^2-a^2}|+c$
Therefore $I_2=\int \large\frac{dx}{\sqrt {x^2-9}}$$=\log |(x)+\sqrt {x^2-9}|+c$
Step 4:
Now $I=I_1+I_2$
$I=3 \sqrt {x^2-9} -\log |(x)+\sqrt {x^2-9}|+c$
answered Apr 19, 2013 by meena.p
 
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