The value of ∫_0^4displaystylex/(√(x^2 + 9)) dx is

The correct answer is Option (4) → 2

Let \(t = x^{2} + 9\). Then

    \( dt = 2x\,dx \)
    \( \displaystyle x\,dx = \frac{1}{2}\,dt \)

The new limits:

    \( x = 0 \Rightarrow t = 9 \)
    \( x = 4 \Rightarrow t = 25 \)

So

    \( \displaystyle \int_{0}^{4}\frac{x}{\sqrt{x^{2} + 9}}\,dx = \frac{1}{2}\int_{9}^{25}t^{-\frac{1}{2}}\,dt \)

    \( \displaystyle = \frac{1}{2}\Big[2\sqrt{t}\Big]_{9}^{25} \)

    \( = 5 - 3 = 2 \)