Step by step solution to "Find the indefinite integral. ∫x2−96xdx"

Albarellak

Answered question

2021-09-01

Find the indefinite integral.
$\int \frac{x^{2} - 96}{x} d x$

Answer & Explanation

Sally Cresswell

Skilled2021-09-02Added 91 answers

Take the integral:
$\int \frac{x^{2} - 96}{x} d x$
For the integrand $\frac{x^{2} - 96}{x}$ , expand out the fraction:
$= \int x - \frac{96}{x} d x$
INTERMEDIATE STEPS:
Expand the following:
$= \int x - \frac{96}{x} d x$
Hint: Express $\frac{x^{2} - 96}{x}$ as a difference of fractions.
$\frac{x^{2} - 96}{x} = \frac{x^{2}}{x} - \frac{96}{x}$ :
$\frac{x^{2}}{x} - \frac{96}{x}$
Hint: For all exponents, $\frac{a^{n}}{a^{m}} = a^{n - m}$ . Apply this to $\frac{x^{2}}{x}$ .
Combine powers. $\frac{x^{2}}{x} = x^{2 - 1}$ :
$x^{2 - 1} - \frac{96}{x}$
Hint: Evaluate $2 - 1$ .
$2 - 1 = 1$ :
Answer: $x - \frac{96}{x}$
Integrate the sum term by term and factor out constants:
$= \int x d x - 96 \int \frac{1}{x} d x$
The integral of x is $\frac{x^{2}}{2}$ :
$= \frac{x^{2}}{2} - 96 \int \frac{1}{x} d x$
The integral of $\frac{1}{x}$ is $\log (x)$ :
Answer:
$= \frac{x^{2}}{2} - 96 \log (x) + constant$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 1

Didn't find what you were looking for?