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Question: If xf(x) = 3(f(x))² + 2, then $\int \frac{2x^2 - 12xf(x)+f(x)}{(6f(x)-x)(x^2-f(x))^2} dx$ equals to...

If xf(x) = 3(f(x))² + 2, then ∫2x2−12xf(x)+f(x)(6f(x)−x)(x2−f(x))2dx\int \frac{2x^2 - 12xf(x)+f(x)}{(6f(x)-x)(x^2-f(x))^2} dx equals to

A

1x2−f(x)+c\frac{1}{x^2-f(x)}+c

B

1x2+f(x)+c\frac{1}{x^2+f(x)}+c

C

1x−f(x)+c\frac{1}{x-f(x)}+c

D

1x+f(x)+c\frac{1}{x+f(x)}+c

Answer

1x2−f(x)+c\frac{1}{x^2-f(x)}+c

Explanation

Solution

To evaluate the integral, we first analyze the given functional equation and then look for a suitable substitution in the integral.

Given functional equation: xf(x)=3(f(x))2+2xf(x) = 3(f(x))^2 + 2

Differentiate both sides with respect to xx: Using the product rule on xf(x)xf(x) and chain rule on 3(f(x))23(f(x))^2: ddx(xf(x))=ddx(3(f(x))2+2)\frac{d}{dx}(xf(x)) = \frac{d}{dx}(3(f(x))^2 + 2) 1⋅f(x)+x⋅f′(x)=3⋅2f(x)⋅f′(x)+01 \cdot f(x) + x \cdot f'(x) = 3 \cdot 2 f(x) \cdot f'(x) + 0 f(x)+xf′(x)=6f(x)f′(x)f(x) + xf'(x) = 6f(x)f'(x)

Rearrange the terms to find f′(x)f'(x): f(x)=6f(x)f′(x)−xf′(x)f(x) = 6f(x)f'(x) - xf'(x) f(x)=(6f(x)−x)f′(x)f(x) = (6f(x) - x)f'(x) So, f′(x)=f(x)6f(x)−xf'(x) = \frac{f(x)}{6f(x) - x}

Now consider the integral: I=∫2x2−12xf(x)+f(x)(6f(x)−x)(x2−f(x))2dxI = \int \frac{2x^2 - 12xf(x)+f(x)}{(6f(x)-x)(x^2-f(x))^2} dx

Let's look at the term (x2−f(x))2(x^2-f(x))^2 in the denominator. This suggests a substitution of the form u=x2−f(x)u = x^2 - f(x).

Let u=x2−f(x)u = x^2 - f(x). Now, find dudu by differentiating uu with respect to xx: du=ddx(x2−f(x))dxdu = \frac{d}{dx}(x^2 - f(x)) dx du=(2x−f′(x))dxdu = (2x - f'(x)) dx

Substitute the expression for f′(x)f'(x) we found earlier: du=(2x−f(x)6f(x)−x)dxdu = \left(2x - \frac{f(x)}{6f(x) - x}\right) dx du=(2x(6f(x)−x)−f(x)6f(x)−x)dxdu = \left(\frac{2x(6f(x) - x) - f(x)}{6f(x) - x}\right) dx du=(12xf(x)−2x2−f(x)6f(x)−x)dxdu = \left(\frac{12xf(x) - 2x^2 - f(x)}{6f(x) - x}\right) dx

Notice that the numerator of the integral is 2x2−12xf(x)+f(x)2x^2 - 12xf(x) + f(x). This is exactly the negative of the numerator in our dudu expression. So, du=−(2x2−12xf(x)+f(x)6f(x)−x)dxdu = -\left(\frac{2x^2 - 12xf(x) + f(x)}{6f(x) - x}\right) dx This implies that 2x2−12xf(x)+f(x)6f(x)−xdx=−du\frac{2x^2 - 12xf(x) + f(x)}{6f(x) - x} dx = -du.

Now, rewrite the integral using our substitution: I=∫1(x2−f(x))2⋅(2x2−12xf(x)+f(x)6f(x)−x)dxI = \int \frac{1}{(x^2-f(x))^2} \cdot \left(\frac{2x^2 - 12xf(x)+f(x)}{6f(x)-x}\right) dx Substitute u=x2−f(x)u = x^2 - f(x) and (2x2−12xf(x)+f(x)6f(x)−x)dx=−du\left(\frac{2x^2 - 12xf(x)+f(x)}{6f(x)-x}\right) dx = -du: I=∫1u2(−du)I = \int \frac{1}{u^2} (-du) I=−∫u−2duI = -\int u^{-2} du

Now, perform the integration: I=−(u−2+1−2+1)+CI = - \left(\frac{u^{-2+1}}{-2+1}\right) + C I=−(u−1−1)+CI = - \left(\frac{u^{-1}}{-1}\right) + C I=1u+CI = \frac{1}{u} + C

Finally, substitute back u=x2−f(x)u = x^2 - f(x): I=1x2−f(x)+CI = \frac{1}{x^2 - f(x)} + C

Comparing this result with the given options, it matches option (A).