Luis's copy of Integration of fraction with power in denominator

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Luis Hernandez 6 years, 4 months ago

Created this as a copy of Integration of fraction with power in denominator.

Name	Status	Author	Last Modified
Integration of fraction with power in denominator	draft	Newcastle University Mathematics and Statistics	20/11/2019 14:50
Luis's copy of Integration of fraction with power in denominator	draft	Luis Hernandez	30/11/2018 18:28
Luis's copy of Integration of fraction with power in denominator	draft	Luis Hernandez	01/01/2019 18:59
Luis's copy of Luis's copy of Integration of fraction with power in denominator	draft	Luis Hernandez	21/12/2018 14:43
Luis's copy of Luis's copy of Luis's copy of Integration of fraction with power in denominator	draft	Luis Hernandez	21/12/2018 14:49

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Question statement

Find the following indefinite integral.

Input all numbers as integers or fractions, not as decimals.

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	4
c	integer	-10
b	integer	-16
d	integer	2
s2	integer	-1
s1	integer	-1
m	integer	-4
n	integer	3
r	integer	-2

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$I=\displaystyle \int \simplify[std]{({b}*x+{c})/(({a}*x+{d})^{n})} dx$

You are given that

$I=\simplify[std]{g(x)*({a}x+{d})^{1-n}}+C$ for a polynomial

$g(x)$ .

You have to find $g(x)$ .

$g(x)=\;$ Unnamed gap

Remember to input all numbers as integers or fractions.

Advice

Let $y = \simplify[std]{{a}*x+{d}}$ .

Then $x=\frac{1}{\var{a}}\simplify[std]{(y-{d})}$ and so we have the numerator $\simplify[std]{{b}*x+{c}}$ becomes in terms of $y$ :

$\simplify[std]{{b}*x+{c} = {b}*1/{a}*(y-{d})+{c}= {m}y+{r}}$ and so

$\simplify[std]{({b}*x+{c})/(({a}*x+{d})^{n})} = \simplify[std]{({m}*y+{r})/(y^{n})={m}/y^{n-1}+{r}/y^{n}}$

Now,

$\int \simplify[std]{({b}x+{c})/({a}*x+{d})^{n}} dx = \int \left(\simplify[std]{{m}/y^{n-1}+{r}/y^{n}} \right)\frac{dx}{dy} dy$

Since $\displaystyle x = \simplify[std]{(y-{d})/{a}}$ then $\displaystyle \frac{dx}{dy} = \frac{1}{\var{a}}$ .

We can now calculate the desired integral:

$\begin{eqnarray*} \int \left(\simplify[std]{{m}/y^{n-1}+{r}/y^{n}}\right) \frac{dx}{dy} dy &=&\frac{1}{\var{a}}\left(\int \simplify[std]{{m}/y^{n-1}}\;dy+\int \simplify[std]{{r}/y^{n}}\;dy \right)\\ &=&\frac{1}{\var{a}}\left(\simplify[std]{{-m}/({n-2}*y^{n-2})+ {-r}/({n-1}*y^{n-1})}\right) + C \\ &=& \simplify[std]{(-{m})/({a*(n-2)}*({a}*x+{d})^{n-2})+(-{r})/({a*(n-1)}*({a}*x+{d})^{n-1}) + C}\\ &=&\simplify[std]{1/({a}x+{d})^{n-1}*(({-m}/{a*(n-2)})*({a}x+{d})-{r}/({a*(n-1)}))}\\ &=&\simplify[std]{1/({a}x+{d})^{n-1}*(({-m}/{(n-2)})*x+{-m*d*(n-1)-r*(n-2)}/{(n-2)*(n-1)*a})} \end{eqnarray*}$
Hence

$g(x)=\simplify[std]{({-m}/{(n-2)})*x+{-m*d*(n-1)-r*(n-2)}/{(n-2)*(n-1)*a}}$

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Luis's copy of Integration of fraction with power in denominator

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Luis Hernandez 6 years, 4 months ago

Error in variable testing condition

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Generated value: `integer`

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Luis's copy of Integration of fraction with power in denominator

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Luis Hernandez 6 years, 4 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`