Find {eq}\displaystyle f {/eq} for the following function.

{eq}\displaystyle f'(x) = \frac{(x+1)}{\sqrt{x}}, \ f(1) = 5 {/eq}

We have the following given data

{eq}\begin{align} f'(x) &= \frac{(x+1)}{\sqrt{x}} \\[0.3cm] f(1) & =5 \\[0.3cm] f (x) &=??\\[0.3cm] \end{align} {/eq}

Solution

We have to integrate the derivative function to find the primitive function or the antiderivative. While integrating, we can take the initial values as the lower limits of integration.

{eq}\begin{align} f(1) & =5 \\[0.3cm] f'(x) &= \frac{(x+1)}{\sqrt{x}} \\[0.3cm] &= \frac{x}{\sqrt{x}}+ \frac{1}{\sqrt{x}}\\[0.3cm] &= \sqrt{x} + \frac{1}{\sqrt{x}}\\[0.3cm] &= x^{\frac{1}{2}} + x^{-\frac{1}{2}} \\[0.3cm] \text{Now,}\int f'(x)dx &= \int \left( x^{\frac{1}{2}} + x^{-\frac{1}{2}} \right) dx & & \left[ \text{Integrate on both sides } \right] \\[0.3cm] \int_{f (1) }^{f (x) } f'(x)dx &= \int_1^x \left( x^{\frac{1}{2}} + x^{-\frac{1}{2}} \right) dx & & \left[ \text{Integrate on both sides. At }~ x=1, f(x)=f(1)=5~ \text{and when }~ x=x, f(x)=f(x) \right] \\[0.3cm] \left[ f(x) \right]_{f (1) }^{f (x) } &= \left[ \frac{x^{\frac{1}{2}+1} }{\frac{1}{2}+1} + \frac{3x^{-\frac{1}{2}+1} }{-\frac{1}{2}+1} \right]_1^x && \left[ \text{Integrate with respect to x on the right-hand side. } \right] \\[0.3cm] f(x)- f (1) &= \left[ \frac{x^{\frac{3}{2}} }{\frac{3}{2}} + \frac{x^{\frac{1}{2}} }{\frac{1}{2}} \right]_1^x \\[0.3cm] f(x)- 5 &= \left[ \frac{2}{3} x^{\frac{3}{2}} + 2 x^{\frac{1}{2}} \right] - \left[ \frac{2}{3} (1)^{\frac{3}{2}} + 2 (1)^{\frac{1}{2}} \right]&& \left[ \text{Apply the limits of integration } \right] \\[0.3cm] f(x) &= \frac{2}{3} x^{\frac{3}{2}} + 2 x^{\frac{1}{2}} -\frac{8}{3} +5 & & \left[ \text{Simplify } \right] \\[0.3cm] f(x) &= \frac{2}{3} x^{\frac{3}{2}} + 2 x^{\frac{1}{2}} + \frac{7}{3} \\[0.3cm] f(x) &= \frac{2}{3} x\sqrt{x} + 2 \sqrt{x} + \frac{7}{3} \\[0.3cm] \end{align} {/eq}

Therefore, {eq}\displaystyle \boxed{\color{blue} { f(x) = \frac{2}{3} x\sqrt{x} + 2 \sqrt{x} + \frac{7}{3} }} {/eq}