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Integrate the function $\frac{x^2-1}{x-1}$ from $3$ to $2$

Step-by-step Solution

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Final Answer

$-\frac{7}{2}$
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Step-by-step Solution

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We could not solve this problem by using the method: Integrals by Partial Fraction Expansion

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Since the upper limit of the integral is less than the lower one, we can rewrite the limits by applying the inverse property of integration limits: If we invert the limits of an integral, it changes sign: $\int_a^bf(x)dx=-\int_b^af(x)dx$

$-\int_{2}^{3}\frac{x^2-1}{x-1}dx$

Learn how to solve definite integrals problems step by step online.

$-\int_{2}^{3}\frac{x^2-1}{x-1}dx$

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Learn how to solve definite integrals problems step by step online. Integrate the function (x^2-1)/(x-1) from 3 to 2. Since the upper limit of the integral is less than the lower one, we can rewrite the limits by applying the inverse property of integration limits: If we invert the limits of an integral, it changes sign: \int_a^bf(x)dx=-\int_b^af(x)dx. Rewrite the expression \frac{x^2-1}{x-1} inside the integral in factored form. Expand the integral \int\left(x+1\right)dx into 2 integrals using the sum rule for integrals, to then solve each integral separately. The integral of a constant is equal to the constant times the integral's variable.

Final Answer

$-\frac{7}{2}$

Exact Numeric Answer

$-3.5$

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Function Plot

Plotting: $\frac{x^2-1}{x-1}$

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1
2
3
4
5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

How to improve your answer:

Main Topic: Definite Integrals

Given a function f(x) and the interval [a,b], the definite integral is equal to the area that is bounded by the graph of f(x), the x-axis and the vertical lines x=a and x=b

Used Formulas

3. See formulas

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