Indefinite Integral of a function $$\int \left(\frac15 x^3 - 2x + \frac3x + e^x \right ) \mathrm dx$$
I came up with
$$F=x^4-x^2+\frac{3x}{\frac12 x^2}+e^x$$
but that was wrong.
 A: \begin{align}
\int \frac{1}{5}(x^3)-2x+\frac{3}{x}+e^x dx & =  \int \frac{1}{5}x^3 dx- \int 2x dx+ \int \frac{3}{x} dx+ \int e^x dx \\ &= \frac{1}{20}x^4 - x^2 + 3 \ln(x) + e^x + C 
\end{align}
Important rules:
•Derivative of $x^n$ is $\frac{1}{n+1}x^{n+1}$ for $n \neq -1$
•Derivative of $x^{-1}$ is $\ln(x)$
A: The first term is $\frac{x^4}{20}$, next $x^2$, next $3 \ln(x)$ the last $\exp(x)$ and a constant
A: The first term and the third terms are wrong.
First term: $\int \frac{1}{5}x^3 dx= \frac{x^4}{20}$. Try differentiating and check $\int x^p dx = \frac{1}{p+1} x^{p+1}$ if $p \neq -1$.
Third term: $\frac{3}{x} = 3x^{-1}$, and $\int x^{-1} = \ln x$, so $\int 3x^{-1} = 3\ln x$.
Finally, don't forget to PLUS $c$!
A: $\int cx^n=c\int x^n=c\frac{x^{n+1}}{n+1}$ when $n\neq -1$
Thus, $\int \frac{1}{5}x^3=\frac{1}{20}x^4$
Also, by the same rule, $\int -2x=-x^2$
$\int \frac{1}{x}=\ln(x)$
Thus, $\int \frac{3}{x}=3\ln(x)$
$\int e^x=e^x$
Thus, by above, the required integral is 
$F(x)=\frac{1}{20}x^4-x^2+3\ln(x)+e^x+c$
Done.
