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? 
 A: $\bf{My \; Solution::}$  $\displaystyle \int (4x-1+3\sqrt{x})dx = 4\int xdx - \int 1 dx+3\int x^{\frac{1}{2}}dx = 4\cdot \frac{x^2}{2}-x+3\cdot \frac{2}{3}x^{\frac{3}{2}}+\mathcal{C}$
Above we have used the formula $\displaystyle \int x^ndx = \frac{x^{n+1}}{n+1}+\mathcal{C}\;,$ Where $n\neq -1$
A: $$\int \left( 4x -1 +3 \sqrt{x}\right) dx$$
$$=4\int x\ dx-\int dx +3\int x^{\frac12} dx$$
$$=\frac{4x^2}{2}-x + \frac{3x^{\frac32}}{\frac32}+C$$
$$=2x^2-x + 2x^{\frac32}+C$$
Here's how we can verify this answer
$$ \frac{d}{dx}\left[ 2x^2-x + 2x^{\frac32}+C \right] $$
$$ =\frac{d}{dx}\left[2x^2\right] -\frac{d}{dx}[x]+ \frac{d}{dx}\left[2x^{\frac32} \right]+\frac{d}{dx}[C]$$
$$ =2\cdot 2x-1+ \frac{2\cdot 3x^{\frac12}}{2} +0$$
$$=4x-1+3x^{\frac12}$$
A: Your answer is not correct. If you integrate term by term, you have: 
\begin{align}
&\int ( 4x -1 +3 \sqrt{x})\mathrm{d}x \\
&\implies \int 4x \mathrm{d}x - \int 1 \mathrm{d}x+\int 3\sqrt{x} \mathrm{d}x \\
&\implies 2x^2 - x + 2x^\frac{3}{2} + C
\end{align}
If you have further questions, let me know!
