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Solve! - Integral Calculus - JEE Main-2

The integral \int \frac{x+2}{(x^{2}+3x+3)\sqrt{x+1}}dxis equal to:

(where C is a constant of integration)

  • Option 1)

    \frac{1}{\sqrt{3}}\cot ^{-1}\left [ \frac{x\sqrt{3}}{\sqrt{x+1}} \right ]+C

  • Option 2)

    \frac{1}{\sqrt{3}}\tan ^{-1}\left [ \frac{x}{\sqrt{3(x+1)}} \right ]+C

  • Option 3)

    \frac{2}{\sqrt{3}}\tan ^{-1}\left [ \frac{x}{\sqrt{3(x+1)}} \right ]+C

  • Option 4)

    \frac{2}{\sqrt{3}}\cot ^{-1}\left [ \frac{x}{\sqrt{x+1}} \right ]+C

 
Answers (1)
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As learnt in concept

Integration by substitution -

The functions when on substitution of the variable of integration to some quantity gives any one of standard formulas.

 

 

- wherein

Since \int f(x)dx=\int f(t)dt=\int f(\theta )d\theta all variables must be converted into single variable ,\left ( t\, or\ \theta \right )

 

 

 \int \frac{(x+2)}{(x^2+3x+3)\sqrt{x+1}}dx

Put \:x+1=t^2; dx=2tdt

I=\int \frac{(t^2 +1)2tdt}{[(t^2-1)^2+3(t^2-1)+3]\sqrt{t^2}}

I=2 \int \frac{(t^2 +1)tdt}{(t^4 + t^2+1)}

I=2 \int \frac{1+1/t^2}{t^2+1+1/t^2} dt

I= \frac{2}{3}\tan^{-1}\frac{(t-1/t)}{\sqrt{3}} +C

I= \frac{2}{\sqrt3}\tan^{-1}\frac{(x)}{\sqrt{3(x+1)}} +C


Option 1)

\frac{1}{\sqrt{3}}\cot ^{-1}\left [ \frac{x\sqrt{3}}{\sqrt{x+1}} \right ]+C

Incorrect

Option 2)

\frac{1}{\sqrt{3}}\tan ^{-1}\left [ \frac{x}{\sqrt{3(x+1)}} \right ]+C

Incorrect

Option 3)

\frac{2}{\sqrt{3}}\tan ^{-1}\left [ \frac{x}{\sqrt{3(x+1)}} \right ]+C

Correct

Option 4)

\frac{2}{\sqrt{3}}\cot ^{-1}\left [ \frac{x}{\sqrt{x+1}} \right ]+C

Incorrect

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