# Find the antiderivative of f(x) = \frac{x}{\sqrt{16 + x^2}}. State the method used.

## Question:

Find the antiderivative of {eq}\; \displaystyle f(x) = \frac{x}{\sqrt{16 + x^2}} {/eq}. {eq}\; {/eq} State the method used.

## Integration By Substitution:

Integration by substitution is a mathematical technique to solve complicated integrals. The substitution method is applied in only those cases in which the given integral is in the form of {eq}\int g(h(x))h'(x) dx {/eq}.

In this method, we substitute {eq}h(x)=t {/eq} and differentiate the substitution and applying this substitution to the integral so that the integral is transformed into another integral which is easier to solve.

Given

{eq}\; \displaystyle f(x) = \dfrac{x}{\sqrt{16 + x^2}} {/eq}.

We have to find the antiderivative of {eq}f(x) {/eq} i.e

{eq}\int f(x) dx=\int \dfrac{x}{\sqrt{16 + x^2}} dx {/eq}

Here, in our problem to solve the antiderivative, we use the substitution method.

We use the substitution {eq}16+x^2=t {/eq} to solve the antiderivative.

Differentiate both sides of the substitution with respect to {eq}x {/eq}

{eq}\begin{align} 16+x^2 &=t\\ 0+2x &=\dfrac{\mathrm{d} t}{\mathrm{d} x}\\ 2xdx &=dt\\ xdx &=\dfrac{dt}{2} \end{align} {/eq}

Applying this substitution to the integral, we have:

{eq}\begin{align} \int \dfrac{x}{\sqrt{16 + x^2}} dx &=\int \dfrac{\dfrac{dt}{2}}{\sqrt{t}}\\ &=\dfrac{1}{2}\int t^{\frac{-1}{2}}dt\\ &=\dfrac{1}{2}\left [ \dfrac{t^{\frac{-1}{2}+1}}{\dfrac{-1}{2}+1} \right ]+C\\ &=\dfrac{1}{2}\left [ \dfrac{t^{\frac{1}{2}}}{\dfrac{1}{2}} \right ]+C\\ &=t^{\frac{1}{2}}+C \end{align} {/eq}

Reversing the integration, we have:

{eq}\color{blue}{\int \dfrac{x}{\sqrt{16 + x^2}} dx=\sqrt{(16+x^2)}+C} {/eq}