Finding Coefficients in Partial Fraction Decomposition

[rudasi]

Hi,

I am an engineering student and I have so far taken Calc 1 & 2 and Ordinary differential equations. I need some online links to question papers. I am looking for some challenging/difficult level questions.

Thanks.

 

[Char. Limit]

You could probably just ask here, and people would be glad to give you questions. I like this one:

$$\int \frac{dx}{1 - x^4}$$

But I'm not sure if you'd consider that challenging.

 

[Nebuchadnezza]

All the fun integration problems I have found

http://www.2shared.com/document/9GniS_-6/Integrals_from_R_to_Z.html

Easy is Calc1
Medium is Calc2
Hard is Real analysis/complex analysis and beyond.

Some of my favs, difficulty varies greatly.

$$\int (1+2x)e^{x^2} \text{dx}$$

$$\int_{-1}^{1} \sqrt{\cos(x)+1} \text{dx}$$

$$\int \frac{1}{\ln(x)} - \frac{1}{\ln(x)^2} \text{dx}$$

$$\int e^{\sqrt{x}} \text{dx}$$

$$\int e^{e^x+x} \text{dx}$$

$$\int_{1}^{e^{\pi}} \sin(\ln(x)) \text{dx}$$

$$\int \sqrt{\frac{x+k}{x}} \text{dx}$$

$$\int \frac{x e^{2x}}{(2x+1)^2} \text{dx}$$

$$\int \frac{ 3^x \cdot 2^x }{ 9^x - 4^x } \text{dx}$$

$$\int_{0}^{\pi/2} \sqrt{\tan(x)} \text{dx}$$

 

[3.1415926535]

All the fun integration problems I have found

http://www.2shared.com/document/9GniS_-6/Integrals_from_R_to_Z.html

Easy is Calc1
Medium is Calc2
Hard is Real analysis/complex analysis and beyond.

Some of my favs, difficulty varies greatly.

$$\int (1+2x)e^{x^2} \text{dx}$$

Are you sure the first one can be written only by using elementary functions? I can't come up with a solution that doesn't include the Error Function
$$Erf(x)=\frac{2}{\sqrt{\pi}}\int e^{-x^2}dx$$

 

[Char. Limit]

Are you sure the first one can be written only by using elementary functions? I can't come up with a solution that doesn't include the Error Function
$$Erf(x)=\frac{2}{\sqrt{\pi}}\int e^{-x^2}dx$$

I believe he intended to write (1+2x²)e^x2, which I believe does have a solution in the elementary functions.

 

[3.1415926535]

I believe he intended to write (1+2x²)e^x2, which I believe does have a solution in the elementary functions.

I believe so too
\begin{array}\\\int (1+2x^2)e^{x^2}dx=\int e^{x^2}dx+\int 2x^2e^{x^2}dx=\int x'e^{x^2}dx+\int 2x^2e^{x^2}dx=xe^{x^2}-\int 2x^2e^{x^2}dx+\int 2x^2e^{x^2}dx=xe^{x^2}\end{array}

 

[Nebuchadnezza]

You are correct, sorry for the confusion =)

 

[3.1415926535]

You are correct, sorry for the confusion =)

I think there is a problem with $$\int \sqrt{\frac{x+k}{x}}dx$$ too
The answer of Wolfram includes the Elliptic Integral of Arcsinhx and the imaginary unit $$i=\sqrt{-1}$$ even when k=1

 

[Nebuchadnezza]

It is not an easy problem by far, but doable. I have the complete solution for it though.
A protip is to change all the constants to n in WolframAlpha, otherwise it thinks you are dealing with a multivariable function.

Here is the correct input.

[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP316419feh8a930dhc4b500003a38eb6h7bb163fh?MSPStoreType=image/gif&s=18&w=470&h=53.jpg

 

[3.1415926535]

It is not an easy problem by far, but doable. I have the complete solution for it though.
A protip is to change all the constants to n in WolframAlpha, otherwise it thinks you are dealing with a multivariable function.

Here is the correct input.

http://www.wolframalpha.com/input/?i=integrate+sqrt((x+n)/x)+dx

Sorry, my mistake. I forgot to enter a parenthesis

 

[dextercioby]

I, too, noticed that

$$\int \sqrt{\frac{x+k}{x}} \, dx$$

is a family of elementary functions.

 

[Nebuchadnezza]

I noticed my answer, and the answer from wolfram alpha varies.

Could anyone check my calculations?

$$\int {\sqrt {\frac{{x + n}}{x}} dx} = \int {u \cdot dx} = \int {u \cdot } \left( { - \frac{{2n \cdot u}}{{{{\left( {{u^2} - 1} \right)}^2}}}du} \right) = - 2n\int {\frac{{{u^2}}}{{{{\left( {{u^2} - 1} \right)}^2}}}du}$$

$${u^2} = \frac{{x + n}}{x} \Leftrightarrow {u^2} \cdot x - x = n \Leftrightarrow x\left( {{u^2} - 1} \right) = n \Leftrightarrow x = \frac{n}{{{u^2} - 1}}$$

$$2u{\rm{ du}} = - \frac{n}{{{x^2}}}dx,dx = - \frac{{2u \cdot {x^2}}}{n}du = - \frac{{2n \cdot u}}{{{{\left( {{u^2} - 1} \right)}^2}}}du$$

$$G\left( u \right) = \frac{{{u^2}}}{{{{\left( {{u^2} - 1} \right)}^2}}} = \left[ {\frac{{{u^2}}}{{{{\left( {u - 1} \right)}^2}{{\left( {u + 1} \right)}^2}}} = \frac{A}{{u - 1}} + \frac{B}{{{{\left( {u - 1} \right)}^2}}} + \frac{C}{{u + 1}} + \frac{D}{{{{\left( {u + 1} \right)}^2}}}} \right]$$

$$\lim_{u \to -1} {\left( {u + 1}\right)^2} G\left( { u} \right) \Rightarrow \left[ {\frac{1}{4} = D} \right] \; \; {\text{ and }} \; \; \lim_{u \to 1} {\left( {u - 1} \right)^2} G\left( u \right) \Rightarrow \left[ {\frac{1}{4} = B} \right]$$

$$\frac{{{u^2}}}{{{{\left( {u - 1} \right)}^2}{{\left( {u + 1} \right)}^2}}} = \frac{A}{{u - 1}} + \frac{1}{{4{{\left( {u - 1} \right)}^2}}} + \frac{C}{{u + 1}} + \frac{1}{{4{{\left( {u + 1} \right)}^2}}}$$

$$G\left( 2 \right) = \left[ {\frac{4}{9} = A + \frac{1}{4} + \frac{1}{3}C + \frac{1}{{36}}} \right],G\left( { - 2} \right) = \left[ {\frac{4}{9} = - \frac{1}{3}A + \frac{1}{4} - C + \frac{1}{{36}}} \right]$$

$$I = \frac{1}{6} = A + \frac{1}{3}C \qquad , \qquad II = \frac{1}{6} = - \frac{1}{3}A - C$$

$$I + 3 \cdot II \Leftrightarrow \frac{1}{6} + \frac{1}{2} = A - A - 3C + \frac{1}{3} \Leftrightarrow \frac{4}{6} = - \frac{8}{3}C \Leftrightarrow C = - \frac{1}{4}$$

$$I \Leftrightarrow \frac{1}{6} = A + \frac{1}{3}\left( { - \frac{1}{4}} \right) \Leftrightarrow A = \frac{2}{{12}} + \frac{1}{{12}} \Leftrightarrow A = \frac{1}{4}$$

$$\frac{{{u^2}}}{{{{\left( {{u^2} - 1} \right)}^2}}} = \frac{1}{{4\left( {u - 1} \right)}} + \frac{1}{{4{{\left( {u - 1} \right)}^2}}} - \frac{1}{{4\left( {u + 1} \right)}} + \frac{1}{{4{{\left( {u + 1} \right)}^2}}}$$

$$\int {\sqrt {\frac{{x + n}}{x}} dx} = - 2n\int {\frac{{{u^2}}}{{{{\left( {{u^2} - 1} \right)}^2}}}du} = - 2n\int {\frac{1}{{4\left( {u - 1} \right)}} + \frac{1}{{4{{\left( {u - 1} \right)}^2}}} - \frac{1}{{4\left( {u + 1} \right)}} + \frac{1}{{4{{\left( {u + 1} \right)}^2}}}du}$$

$$\int {\frac{1}{{4{{\left( {u - 1} \right)}^2}}}} du = \frac{1}{4}\int {\frac{1}{{{{\left( k \right)}^2}}}dk = \frac{1}{4}\left( { - \frac{1}{k}} \right)} = - \frac{1}{4}\frac{1}{{\left( {u - 1} \right)}} + R$$

$$\int {\frac{1}{{4{{\left( {u + 1} \right)}^2}}}} du = \frac{1}{4}\int {\frac{1}{{{{\left( v \right)}^2}}}dv = \frac{1}{4}\left( { - \frac{1}{v}} \right)} = - \frac{1}{4}\frac{1}{{\left( {u + 1} \right)}} + T$$


$$\int {\sqrt {\frac{{x + n}}{x}} dx} = - 2n\left[ {\frac{1}{4}\ln \left| {u - 1} \right| - \frac{1}{4}\frac{1}{{\left( {u - 1} \right)}} - \frac{1}{4}\ln \left| {u + 1} \right| - \frac{1}{4}\frac{1}{{\left( {u + 1} \right)}}} \right]$$

$$\int {\sqrt {\frac{{x + n}}{x}} dx} = \frac{1}{2}n\left[ { - \ln \left| {u - 1} \right| + \frac{1}{{u - 1}} + \ln \left| {u + 1} \right| + \frac{1}{{u + 1}}} \right]$$

$$\int {\sqrt {\frac{{x + n}}{x}} dx} = \frac{1}{2}n\left[ {\frac{{2u}}{{{u^2} - 1}} + \ln \left| {\frac{{u + 1}}{{u - 1}}} \right|} \right] + W \qquad \text{where} \qquad u = \sqrt {\frac{{x + n}}{x}}$$

$$\underline{\underline {\int {\sqrt {\frac{{x + n}}{x}} dx} = x\sqrt {\frac{{x + n}}{x}} + \frac{1}{2}n\ln \left| { - \frac{2}{n}x\left( {1 + \sqrt {\frac{{x + n}}{x}} } \right) - 1} \right| + W}}$$

 

[3.1415926535]

I noticed my answer, and the answer from wolfram alpha varies.

Could anyone check my calculations?


$$A=-1/4$$ and $$C=1/4$$, not the other way around(I think)

\begin{array}\\\frac{u^2}{(u-1)^2(u+1)^2}=\frac{A}{u+1}+\frac{B}{(u+1)^2}+\frac{C}{u-1}+\frac{D}{(u-1)^2}\Leftrightarrow u^2=(Au+A)(u-1)^2+B(u-1)^2+(Cu-C)(u+1)^2+D(u+1)^2\Leftrightarrow\\
(Au+A)(u^2-2u+1)+B(u^2-2u+1)+(Cu-C)(u^2+2u+1)+D(u^2+2u+1)-u^2=0\Leftrightarrow\\
(Au^3-2Au^2+Au+Au^2-2Au+A)+(Bu^2-2Bu+B)+(Cu^3+2Cu^2+Cu-Cu^2-2Cu-C)+(Du^2+2Du+D)-u^2=0\Leftrightarrow\\
u^3(A+C)+u^2(-A+B+C+D-1)+u(-A-2B-C+2D)+(A+B-C+D)=0\\
\begin{Bmatrix}
A& +& C& &=0 \\
-A& +B& +C& +D&=1 \\
-A& -2B& -C& 2D&=0 \\
A& +B & -C & +D&=0
\end{Bmatrix}\Leftrightarrow
\begin{Bmatrix}
A& =-C& & & \\
-A& +B& -A& +D&=1 \\
-A& -2B& +A& +2D&=0 \\
A& +B & +A & +D&=0
\end{Bmatrix}\Leftrightarrow
\begin{Bmatrix}
A& =-C& & & \\
-2A& +B & &+D &=1 \\
& -2B& & +2D&=0 \\
2A& +B & & +D&=0
\end{Bmatrix}\\
\begin{Bmatrix}
A& =-C& & \\
-2A& +2B & &=1 \\
B& =D& & \\
2A& +2B & &=0
\end{Bmatrix}\Rightarrow 4B=1\Leftrightarrow B=\frac{1}{4}=D\\
\begin{Bmatrix}
A& =-C& & \\
B& =D& =\frac{1}{4}& \\
2A& +\frac{1}{2} & &=0
\end{Bmatrix}\Rightarrow A=-\frac{1}{4}, C=\frac{1}{4}\end{array}