The value of ∫_1^3(x^2)/(x^2 + (4 - x)^2) dx is

The correct answer is Option (2) → 1

Let

    \( \displaystyle I = \int_{1}^{3}\frac{x^{2}}{x^{2} + (4 - x)^{2}}\,dx \)

Here a + b = 1 + 3 = 4. By the property \(\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx\), replace x by 4 − x:

    \( \displaystyle I = \int_{1}^{3}\frac{(4 - x)^{2}}{(4 - x)^{2} + x^{2}}\,dx \)

Adding the two forms of I, the numerators add up to the denominator:

    \( \displaystyle 2I = \int_{1}^{3}\frac{x^{2} + (4 - x)^{2}}{x^{2} + (4 - x)^{2}}\,dx \)

    \( \displaystyle = \int_{1}^{3}1\,dx = 2 \)

So

    \( I = 1 \)