Integration help

checkcheek

where do i start with this, or what direction do i go to make it ready for integration


integrate 5x+1/(x+3)^2 dx

any help would be great

ClovisI

Try the substitution u=x+3.

TheBody

checkcheek wrote: »

where do i start with this, or what direction do i go to make it ready for integration


integrate 5x+1/(x+3)^2 dx

any help would be great

Partial fractions is the way to go in this case. The idea is to break the problem into a number of simplier integrals.

Your problem is to:

[latex]\int {\frac{5x+1}{(x+3)^2}dx} [/latex]

Begin by considering equation:

[latex]\frac{5x+1}{(x+3)^2} =\frac{A}{x+3}+\frac{B}{(x+3)^2}[/latex]

We need to solve for A and B. Adding the two fractions on the right gives:

[latex]\frac{5x+1}{(x+3)^2} =\frac{A(x+3)+B}{(x+3)^2}[/latex]

Next, by comparison:

[latex]5x+1=A(x+3)+B[/latex]

Multiplying out gives:

[latex]5x+1=Ax+(3A+B)[/latex]

Now, if we compare coefficients on both sides:

[latex]5x=Ax[/latex]

So therefore [latex]A=5[/latex].

We also have [latex]1=(3A+B)[/latex].

But since we know A=5 we have [latex]1=(15+B)[/latex]

so solving gives [latex]B=-14[/latex]

We can now return to the problem of integration. Since we now know that A and B are we can re-write our initial integral as:

[latex]\int {\frac{5x+1}{(x+3)^2}dx}=\int{\frac{5}{x+3}}dx-\int{\frac{14}{(x+3)^2}}dx [/latex]

[latex]=5\int{\frac{1}{x+3}}dx-14\int{\frac{1}{(x+3)^2}}dx [/latex]

Can you finish it from there op?

checkcheek

TheBody wrote: »

Partial fractions is the way to go in this case. The idea is to break the problem into a number of simplier integrals.

Your problem is to:

[latex]\int {\frac{5x+1}{(x+3)^2}dx} [/latex]

Begin by considering equation:

[latex]\frac{5x+1}{(x+3)^2} =\frac{A}{x+3}+\frac{B}{(x+3)^2}[/latex]

We need to solve for A and B. Adding the two fractions on the right gives:

[latex]\frac{5x+1}{(x+3)^2} =\frac{A(x+3)+B}{(x+3)^2}[/latex]

Next, by comparison:

[latex]5x+1=A(x+3)+B[/latex]

Multiplying out gives:

[latex]5x+1=Ax+(3A+B)[/latex]

Now, if we compare coefficients on both sides:

[latex]5x=Ax[/latex]

So therefore [latex]A=5[/latex].

We also have [latex]1=(3A+B)[/latex].

But since we know A=5 we have [latex]1=(15+B)[/latex]

so solving gives [latex]B=-14[/latex]

We can now return to the problem of integration. Since we now know that A and B are we can re-write our initial integral as:

[latex]\int {\frac{5x+1}{(x+3)^2}dx}=\int{\frac{5}{x+3}}dx-\int{\frac{14}{(x+3)^2}}dx [/latex]

[latex]=5\int{\frac{1}{x+3}}dx-14\int{\frac{1}{(x+3)^2}}dx [/latex]

Can you finish it from there op?

Thats great thank you so so so much

checkcheek

Sorry, Ive another question on a completly different topic

I dont know where the root 3 comes from, i understand the rest but i feel you need to know how to get the root 3

Same where it has X1, X2, X3 where it has V1 over root 2.

Sorry for all the question. But its so much help.

Thank you, thank you, thank you

TheBody

checkcheek wrote: »

Sorry, Ive another question on a completly different topic

I dont know where the root 3 comes from, i understand the rest but i feel you need to know how to get the root 3

Same where it has X1, X2, X3 where it has V1 over root 2.

Sorry for all the question. But its so much help.

Thank you, thank you, thank you

All that's happening there is scaling the vector so that is has a length of 1.

For example, in the first case it starts with the vector [latex]w_3=(1,1,1) [/latex]. As [latex]x_3=\frac{w_3}{||w_3|| } [/latex] we need to find the length of [latex]w_3 [/latex].

[latex]||w_3||=\sqrt{1^2+1^2+1^2}=\sqrt{3}[/latex].

Therefore [latex]x_3=(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} )[/latex].

Exactly the same thing is happening in the second question you asked.

checkcheek

TheBody wrote: »

All that's happening there is scaling the vector so that is has a length of 1.

For example, in the first case it starts with the vector [latex]w_3=(1,1,1) [/latex]. As [latex]x_3=\frac{w_3}{||w_3|| } [/latex] we need to find the length of [latex]w_3 [/latex].

[latex]||w_3||=\sqrt{1^2+1^2+1^2}=\sqrt{3}[/latex].

Therefore [latex]x_3=(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} )[/latex].

Exactly the same thing is happening in the second question you asked.

You are an absolute God send, thank you so much.

TheBody

checkcheek wrote: »

You are an absolute God send, thank you so much.

No problem. Glad I could help.

Michael Collins

TheBody wrote: »

Partial fractions is the way to go in this case. The idea is to break the problem into a number of simplier integrals.

Your problem is to:

[latex]\int {\frac{5x+1}{(x+3)^2}dx} [/latex]

Begin by considering equation:

[latex]\frac{5x+1}{(x+3)^2} =\frac{A}{x+3}+\frac{B}{(x+3)^2}[/latex]

We need to solve for A and B. Adding the two fractions on the right gives:

[latex]\frac{5x+1}{(x+3)^2} =\frac{A(x+3)+B}{(x+3)^2}[/latex]

Next, by comparison:

[latex]5x+1=A(x+3)+B[/latex]

Multiplying out gives:

[latex]5x+1=Ax+(3A+B)[/latex]

Now, if we compare coefficients on both sides:

[latex]5x=Ax[/latex]

So therefore [latex]A=5[/latex].

We also have [latex]1=(3A+B)[/latex].

But since we know A=5 we have [latex]1=(15+B)[/latex]

so solving gives [latex]B=-14[/latex]

We can now return to the problem of integration. Since we now know that A and B are we can re-write our initial integral as:

[latex]\int {\frac{5x+1}{(x+3)^2}dx}=\int{\frac{5}{x+3}}dx-\int{\frac{14}{(x+3)^2}}dx [/latex]

[latex]=5\int{\frac{1}{x+3}}dx-14\int{\frac{1}{(x+3)^2}}dx [/latex]

Can you finish it from there op?

I guess it's down to whatever you're comfortable with but I'd prefer to use ClovisI's idea of the substitution u = x+3 (=> x = u-3 and dx = du) for this problem and avoid the partial fraction expansion.