Find the integral: $\int ( 4x -1 +3 \sqrt{x})\mathrm{d}x$ I have to find the following integral: 

$$\int ( 4x -1 +3 \sqrt{x})\mathrm{d}x$$

My answer is $2x^2 -\ 1x + \frac{2\sqrt{27}}{3}$. Am I right?
$\bf{My \; Solution::}$  $\displaystyle \int (4x-1+3\sqrt{x})dx = 4\int xdx - \int 1 dx+3\int x^{\frac{1}{2}}dx = 4\cdot \frac{x^2}{2}-x+3\cdot \frac{2}{3}x^{\frac{3}{2}}+\mathcal{C}$

Above we have used the formula $\displaystyle \int x^ndx = \frac{x^{n+1}}{n+1}+\mathcal{C}\;,$ Where $n\neq -1$
Your answer is not correct. If you integrate term by term, you have: 

\begin{align}
&amp;\int ( 4x -1 +3 \sqrt{x})\mathrm{d}x \\
&amp;\implies \int 4x \mathrm{d}x - \int 1 \mathrm{d}x+\int 3\sqrt{x} \mathrm{d}x \\
&amp;\implies 2x^2 - x + 2x^\frac{3}{2} + C
\end{align}

If you have further questions, let me know!