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Let’s check out your problem:
Evaluate
∫
3
4
4
x
2
−
9
x
+
1
x
−
2
d
x
\int_{3}^{4} \frac{4 x^{2}-9 x+1}{x-2} d x
∫
3
4
x
−
2
4
x
2
−
9
x
+
1
d
x
. Write your answer in simplest form with all logs condensed into a single logarithm (if necessary).
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Math Problems
Calculus
Find indefinite integrals using the substitution
Full solution
Q.
Evaluate
∫
3
4
4
x
2
−
9
x
+
1
x
−
2
d
x
\int_{3}^{4} \frac{4 x^{2}-9 x+1}{x-2} d x
∫
3
4
x
−
2
4
x
2
−
9
x
+
1
d
x
. Write your answer in simplest form with all logs condensed into a single logarithm (if necessary).
Perform Polynomial Long Division:
We will first perform polynomial
long division
to simplify the integrand
(
4
x
2
−
9
x
+
1
)
/
(
x
−
2
)
(4x^2 - 9x + 1) / (x - 2)
(
4
x
2
−
9
x
+
1
)
/
(
x
−
2
)
.
Simplify Integrand:
Performing the division, we get:
\newline
(
4
x
2
−
9
x
+
1
)
:
(
x
−
2
)
=
4
x
+
1
(4x^2 - 9x + 1) : (x - 2) = 4x + 1
(
4
x
2
−
9
x
+
1
)
:
(
x
−
2
)
=
4
x
+
1
with a remainder of
−
1
-1
−
1
.
\newline
So,
(
4
x
2
−
9
x
+
1
)
/
(
x
−
2
)
=
4
x
+
1
−
1
/
(
x
−
2
)
(4x^2 - 9x + 1) / (x - 2) = 4x + 1 - 1/(x - 2)
(
4
x
2
−
9
x
+
1
)
/
(
x
−
2
)
=
4
x
+
1
−
1/
(
x
−
2
)
.
Write Integral as Sum:
Now we can write the integral as the sum of two simpler integrals:
\newline
∫
x
=
3
x
=
4
(
4
x
+
1
−
1
x
−
2
)
d
x
\int_{x = 3}^{x = 4} (4x + 1 - \frac{1}{x - 2}) \, dx
∫
x
=
3
x
=
4
(
4
x
+
1
−
x
−
2
1
)
d
x
.
Integrate Each Term:
We will integrate each term separately:
\newline
∫
(
4
x
+
1
)
d
x
=
2
x
2
+
x
+
C
\int(4x + 1) \, dx = 2x^2 + x + C
∫
(
4
x
+
1
)
d
x
=
2
x
2
+
x
+
C
\newline
and
\newline
∫
(
−
1
/
(
x
−
2
)
)
d
x
=
−
ln
∣
x
−
2
∣
+
C
\int(-1/(x - 2)) \, dx = -\ln|x - 2| + C
∫
(
−
1/
(
x
−
2
))
d
x
=
−
ln
∣
x
−
2∣
+
C
.
Evaluate Definite Integral:
Now we evaluate the definite integral from
x
=
3
x = 3
x
=
3
to
x
=
4
x = 4
x
=
4
:
\newline
∫
3
4
(
4
x
+
1
−
1
x
−
2
)
d
x
=
[
2
x
2
+
x
−
ln
∣
x
−
2
∣
]
3
4
\int_{3}^{4}(4x + 1 - \frac{1}{x - 2}) \, dx = [2x^2 + x - \ln|x - 2|]_{3}^{4}
∫
3
4
(
4
x
+
1
−
x
−
2
1
)
d
x
=
[
2
x
2
+
x
−
ln
∣
x
−
2∣
]
3
4
.
Upper Limit Calculation:
Plugging in the upper limit
x
=
4
x = 4
x
=
4
:
2
(
4
)
2
+
4
−
ln
∣
4
−
2
∣
=
2
(
16
)
+
4
−
ln
∣
2
∣
=
32
+
4
−
ln
(
2
)
2(4)^2 + 4 - \ln|4 - 2| = 2(16) + 4 - \ln|2| = 32 + 4 - \ln(2)
2
(
4
)
2
+
4
−
ln
∣4
−
2∣
=
2
(
16
)
+
4
−
ln
∣2∣
=
32
+
4
−
ln
(
2
)
.
Lower Limit Calculation:
Plugging in the lower limit
x
=
3
x = 3
x
=
3
:
2
(
3
)
2
+
3
−
ln
∣
3
−
2
∣
=
2
(
9
)
+
3
−
ln
∣
1
∣
=
18
+
3
−
ln
(
1
)
=
18
+
3
−
0
=
21
2(3)^2 + 3 - \ln|3 - 2| = 2(9) + 3 - \ln|1| = 18 + 3 - \ln(1) = 18 + 3 - 0 = 21
2
(
3
)
2
+
3
−
ln
∣3
−
2∣
=
2
(
9
)
+
3
−
ln
∣1∣
=
18
+
3
−
ln
(
1
)
=
18
+
3
−
0
=
21
.
Subtract Values:
Subtract the value at the lower limit from the value at the upper limit:
\newline
(
32
+
4
−
ln
(
2
)
)
−
(
21
)
=
36
−
ln
(
2
)
−
21
=
15
−
ln
(
2
)
(32 + 4 - \ln(2)) - (21) = 36 - \ln(2) - 21 = 15 - \ln(2)
(
32
+
4
−
ln
(
2
))
−
(
21
)
=
36
−
ln
(
2
)
−
21
=
15
−
ln
(
2
)
.
Final Answer:
The final answer in simplest form with all logs condensed into a single logarithm is:
\newline
15
−
ln
(
2
)
15 - \ln(2)
15
−
ln
(
2
)
.
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\newline
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x
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x
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∫
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x
2
5
3
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x
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6
x
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x
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\newline
∫
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x
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x
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∫
6
x
2
2
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x
d
x
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\newline
∫
−
3
x
sin
(
−
2
x
)
d
x
\int-3 x \sin (-2 x) d x
∫
−
3
x
sin
(
−
2
x
)
d
x
\newline
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