The integral $\int_{2}^{4} \frac{\log \mathrm{x}^{2}}{\log \mathrm{x}^{2}+\log \left(36-12 \mathrm{x}+\mathrm{x}^{2}\right)} \mathrm{dx}$ is equal to :

JEE Main · 2015 · 04 Apr 2015 · Offline · Definite Integrals

The integral

\int_{2}^{4} \frac{l o g x ^{2}}{l o g x ^{2} + l o g ( 36 - 12 x + x ^{2} )} dx

is equal to :

Answer: (B)

Spot that

36 - 12 x + x^{2} = (x - 6)^{2} = (6 - x)^{2}

, so the integrand is

\frac{l o g x ^{2}}{l o g x ^{2} + l o g ( 6 - x ) ^{2}} = \frac{l o g x}{l o g x + l o g ( 6 - x )}

. Call the integral

I

and apply the king's-property substitution

x \mapsto 6 - x

, which keeps the limits

2

4

but swaps

x

and

6 - x

, turning the integrand into

\frac{l o g ( 6 - x )}{l o g ( 6 - x ) + l o g x}

. Adding the two forms, the integrands sum to

1

, so

2 I = \int_{2}^{4} 1 d x = 2

, giving

I = 1

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option

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