Simplify and integrate the following: integral from [((x2 - 5/x) × x^(−1/3) + 2sec2 x + 5 sin 3x − 6/x)] dx

Question

ScanSolve AI Tutor · Accepted Answer

The problem asks to simplify and evaluate the indefinite integral:

\int \left( \left( x^2 - \frac{5}{x} \right) x^{-1/3} + 2 \sec^2 x + 5 \sin 3x - \frac{6}{x} \right) dx

Simplify the integrand by distributing $x^{-1/3}$ and combining terms.
$\left( x^2 - \frac{5}{x} \right) x^{-1/3} = x^2 \cdot x^{-1/3} - \frac{5}{x} \cdot x^{-1/3}$
Using the exponent rule $a^m \cdot a^n = a^{m+n}$ :
$x^{2 - 1/3} - 5 x^{-1 - 1/3} = x^{6/3 - 1/3} - 5 x^{-3/3 - 1/3} = x^{5/3} - 5 x^{-4/3}$
So the integral becomes:
$\int \left( x^{5/3} - 5 x^{-4/3} + 2 \sec^2 x + 5 \sin 3x - \frac{6}{x} \right) dx$
Integrate each term separately using the standard integration rules
- $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$ )
- $\int \frac{1}{x} dx = \ln|x| + C$
- $\int \sec^2 x dx = \tan x + C$
- $\int \sin(ax) dx = -\frac{1}{a} \cos(ax) + C$
Apply the integration rules to each term.
For $x^{5/3}$ :
$\int x^{5/3} dx = \frac{x^{5/3+1}}{5/3+1} = \frac{x^{8/3}}{8/3} = \frac{3}{8} x^{8/3}$
For $-5 x^{-4/3}$ :
$\int -5 x^{-4/3} dx = -5 \frac{x^{-4/3+1}}{-4/3+1} = -5 \frac{x^{-1/3}}{-1/3} = 15 x^{-1/3}$
For $2 \sec^2 x$ :
$\int 2 \sec^2 x dx = 2 \tan x$
For $5 \sin 3x$ :
$\int 5 \sin 3x dx = 5 \left( -\frac{1}{3} \cos 3x \right) = -\frac{5}{3} \cos 3x$
For $-\frac{6}{x}$ :
$\int -\frac{6}{x} dx = -6 \ln|x|$
Combine all the integrated terms and add the constant of integration, $C$ .
$\frac{3}{8} x^{8/3} + 15 x^{-1/3} + 2 \tan x - \frac{5}{3} \cos 3x - 6 \ln|x| + C$
The final answer is:
$\boxed{\frac{3}{8} x^{8/3} + 15 x^{-1/3} + 2 \tan x - \frac{5}{3} \cos 3x - 6 \ln|x| + C}$

Simplify and integrate the following: integral from [((x2 - 5/x) * x^(-1/3) + 2sec2 x + 5 sin 3x - 6/x)] dx