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Indefinite Integrals question

2021 · 27 Aug · Shift 1 · Q37
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Indefinite Integrals question

2021 · 27 Aug · Shift 1 · Q37

JEE MainMathematicsIndefinite IntegralsNumerical+4 / −1
If ∫dx(x2+x+1)2=atan⁡−1(2x+13)+b(2x+1x2+x+1)+C\int {{{dx} \over {{{({x^2} + x + 1)}^2}}} = a{{\tan }^{ - 1}}\left( {{{2x + 1} \over {\sqrt 3 }}} \right) + b\left( {{{2x + 1} \over {{x^2} + x + 1}}} \right) + C}∫(x2+x+1)2dx​=atan−1(3​2x+1​)+b(x2+x+12x+1​)+C, x > 0 where C is the constant of integration, then the value of 9(3a+b)9\left( {\sqrt 3 a + b} \right)9(3​a+b) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 15

  1. We need to evaluate
∫dx(x2+x+1)2.\int \frac{dx}{(x^2+x+1)^2}.∫(x2+x+1)2dx​.

We are given that

We must find 9(3a+b)9(\sqrt3 a+b)9(3​a+b).

  1. First complete the square in the quadratic:

Also, let

Now,

Therefore,

So the integral becomes

=∫du2(u2+3)216=8∫du(u2+3)2.=\int \frac{\frac{du}{2}}{\frac{(u^2+3)^2}{16}} =8\int \frac{du}{(u^2+3)^2}.=∫16(u2+3)2​2du​​=8∫(u2+3)2du​.
  1. Now use the standard result
=u2a2(u2+a2)+12a3tan⁡−1(ua)+C.=\frac{u}{2a^2(u^2+a^2)}+\frac{1}{2a^3}\tan^{-1}\left(\frac{u}{a}\right)+C.=2a2(u2+a2)u​+2a31​tan−1(au​)+C.

Here a2=3a^2=3a2=3, so a=3a=\sqrt3a=3​. Hence

=u6(u2+3)+163tan⁡−1(u3)+C.=\frac{u}{6(u^2+3)}+\frac{1}{6\sqrt3}\tan^{-1}\left(\frac{u}{\sqrt3}\right)+C.=6(u2+3)u​+63​1​tan−1(3​u​)+C.

Multiplying by 888,

So,

  1. Substitute back u=2x+1u=2x+1u=2x+1 and note that

Thus,

Therefore,

Comparing with

we get

  1. Now compute

Hence,

  1. Final answer: 15\boxed{15}15​

This matches the stored correct answer.

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