Verifying an indefinite integral solution

Question

I'm working on indefinite integrals right now. I'm given the problem

$$\int(\frac{8}{x}-\frac{5}{x^2}+\frac{6}{x^3}) dx$$

My worksheet gives the answer $8\ln(x)+\frac{5}{x}-\frac{3}{x^2}+c$

I'm trying to figure out if my answer is correct in a different form. So I tried making my solution equal to the given solution to test it, but so far I don't think that has worked. Here's how I approached this problem:

\begin{align*} \int\frac{8}{x}dx&=8\int\frac{1}{x}dx \implies 8\ln(|x|) \\ \int-\frac{5}{x^2}dx&=-5\int\frac{1}{x}dx \implies -5\ln(|x^2|) \\ \int\frac{6}{x^3}dx&=6\int\frac{1}{x^3}dx \implies6\ln(|x^3|) \\ &= 8\ln(x)-5\ln(x^2)+6\ln(x^3) \\ &= \ln(x^8)-\ln(x^2)^5+\ln(x^3)^{6} \\ &= \frac{\ln(x)^8}{\ln(x)^{10}}+\ln(x)^{18} \\ &= \ln(\frac{1}{x^2}+\ln(x)^{18}) \\ &= -2\ln(x)+18\ln(x) \\ &= 16\ln(x)+c \end{align*}

Answer 1

Careful!

The primitive of $x^a$ is $\frac{x^{a+1}}{a+1}$ for all $a\neq -1$.

Only for $a=-1$ we have that the primitive is $\ln(|x|)$.

Answer 2

You have to use this:

$\int x^{n} dx = \frac{x^{n+1}}{n+1} +c \; \: \: \: \mathit{where \: n \neq -1}$

The full integral would be like this:

$ 8\int \frac{1}{x} dx = 8\ln |x| + c $

$ 5\int x^{-2} dx = 5*\frac{x^{-2+1}}{-2+1} = \frac{-5}{x} + c $

$ 6\int x^{-3} dx = 6*\frac{x^{-3+1}}{-3+1} = \frac{-3}{x^{2}} + c $

All together:

$8\ln |x| + \frac{5}{x} - \frac{3}{x^{2}} + c$

Answer 3

Your answer is incorrect because $$\int -\frac{5}{x^2} \, dx \neq \int -\frac{5}{x} \, dx$$just like $$\int\frac{6}{x^3} \, dx \neq \int \frac{6}{x} \, dx.$$ You have to use the power rule $\int x^a \, dx = \frac{x^{a+1}}{a+1}.$ The reason why $\int \frac{1}{x} \, dx = \ln{x}:$

An integral of a function $f(x)$ is such that the derivative of the integral is the function $f(x)$. The derivative of $\ln{x}$ is derived using implicit differentiation: $$y=\ln{x}$$ $$e^y=x$$Now if we take the derivative: $$e^y*\frac{dy}{dx}=1$$ Because $y=\ln{x}$: $$x\frac{dy}{dx}=1$$ $$\frac{dy}{dx}=\frac{1}{x} \text{QED}$$