∫25x2+49−14xx2−14x+49+x2dx=

∫25[x2+49−14x][x2−14x+49]+[x2]dx=∫25[(7−x)2][(7−x)2]+[x2]dx=∫05[(2+5−x)2][(2+5−x)2]+[x2]dx∴∫abf(a+b−x)f(x)+f(a+b−xdx=b−a2here f(x)=[x2]∴I=5−22=3/2

∫25x2+49−14xx2−14x+49+x2dx=

∫25[x2+49−14x][x2−14x+49]+[x2]dx=∫25[(7−x)2][(7−x)2]+[x2]dx=∫05[(2+5−x)2][(2+5−x)2]+[x2]dx∴∫abf(a+b−x)f(x)+f(a+b−xdx=b−a2here f(x)=[x2]∴I=5−22=3/2

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$\int_{2}^{5} \frac{[x^{2} + 49 - 14 x]}{[x^{2} - 14 x + 49] + [x^{2}]} d x =$

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$\frac{5}{2}$

$\frac{7}{2}$

$\frac{1}{2}$

$\frac{3}{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} \int_{2}^{5} \frac{[x^{2} + 49 - 14 x]}{[x^{2} - 14 x + 49] + [x^{2}]} d x \\ = \int_{2}^{5} \frac{[{(7 - x)}^{2}]}{[{(7 - x)}^{2}] + [x^{2}]} d x \\ = \int_{0}^{5} \frac{[{(2 + 5 - x)}^{2}]}{[{(2 + 5 - x)}^{2}] + [x^{2}]} d x \\ [∴ \int_{a}^{b} \frac{f (a + b - x)}{f (x) + f (a + b - x} d x = \frac{b - a}{2}] \\ h e r e f (x) = [x^{2}] \\ ∴ I = \frac{5 - 2}{2} = 3 / 2 \end{array}$