Integration help

Re: Integral of 3x/(Sqrt[4x+20])?

∫3xdx4x+20 = 32∫xdxx+5\displaystyle \int\frac{3xdx}{\sqrt{4x+20}} \ = \ \frac{3}{2}\int\frac{xdx}{\sqrt{x+5}}

Now, let u = x+5, du = dx2u   ⟹   dx = 2udu.\displaystyle Now, \ let \ u \ = \ \sqrt{x+5}, \ du \ = \ \frac{dx}{2u} \ \implies \ dx \ = \ 2udu.

u = x+5,   ⟹   x = u2−5.\displaystyle u \ = \ \sqrt{x+5}, \ \implies \ x \ = \ u^2-5.

Hence, we have: 32∫(u2−5)2uduu = 3∫(u2−5)du = 3[(u3/3)−5u] + C,\displaystyle Hence, \ we \ have: \ \frac{3}{2}\int\frac{(u^2-5)2udu}{u} \ = \ 3\int(u^2-5)du \ = \ 3[(u^3/3)-5u] \ + \ C,

= u3−15u + C = (sub) (x+5)3/2−15(x+5)1/2 + C = (x+5)1/2[x+5−15]+ C,\displaystyle = \ u^3-15u \ + \ C \ = \ (sub) \ (x+5)^{3/2}-15(x+5)^{1/2} \ + \ C \ = \ (x+5)^{1/2}[x+5-15]+ \ C,

= (x+5)1/2(x−10)+C, I′ll leave the check up to you.\displaystyle = \ (x+5)^{1/2}(x-10)+C, \ I'll \ leave \ the \ check \ up \ to \ you.
 
Re: Integral of 3x/(Sqrt[4x+20])?

How'd you make the move of factoring out a 2 from Sqrt[4x+20]???
 
Re: Integral of 3x/(Sqrt[4x+20])?

4x+20 = 4(x+5) = 4x+5 = 2x+5\displaystyle \sqrt{4x+20} \ = \ \sqrt{4(x+5)} \ = \ \sqrt4 \sqrt{x+5} \ = \ 2\sqrt{x+5}

If you did not see this, then what can I say?\displaystyle If \ you \ did \ not \ see \ this, \ then \ what \ can \ I \ say?
 
Re: Integral of 3x/(Sqrt[4x+20])?

BigGlenntheHeavy said:
4x+20 = 4(x+5) = 4x+5 = 2x+5\displaystyle \sqrt{4x+20} \ = \ \sqrt{4(x+5)} \ = \ \sqrt4 \sqrt{x+5} \ = \ 2\sqrt{x+5}

If you did not see this, then what can I say?\displaystyle If \ you \ did \ not \ see \ this, \ then \ what \ can \ I \ say?
Click to expand...
Ohhhh I see.

One last question, if I take the anti-derivative of this one more time, it's basically just expanding the factor and doing power rule, right?

And end up with Cx, right?
 
Re: Integral of 3x/(Sqrt[4x+20])?

Look, follow what I did above, as I′m not sure what you are talking about.\displaystyle Look, \ follow \ what \ I \ did \ above, \ as \ I'm \ not \ sure \ what \ you \ are \ talking \ about.
 
Re: Integral of 3x/(Sqrt[4x+20])?

\(\displaystyle \If\ I \ wanted \ to \ take \ the \ integral \ of \int\sqrt{x+5}\\(x-10)\+\C\ one \ more \ time, \ what\ would \ it \ look \ like?\\)

? ? (x+5)*(x-10) , u = x-10, du=dx

?? u+C

=

I get 2(x-10)^(1/2) + Cx
 
Re: Integral of 3x/(Sqrt[4x+20])?

∫(x+5) (x−10)dx = 2[(x+5)3/2(x−20)]5 + C\displaystyle \int\sqrt{(x+5)} \ (x-10)dx \ = \ \frac{2[(x+5)^{3/2}(x-20)]}{5} \ + \ C

Use basically the same strategy.\displaystyle Use \ basically \ the \ same \ strategy.
 
Re: Integral of 3x/(Sqrt[4x+20])?

What u-substitution did u use there? x+5? then factored out the x-10? How'd you get x-20 ? Sorry the algebra is confusing me somewhat...
 
Re: Integral of 3x/(Sqrt[4x+20])?

\(\displaystyle OK \ Jaskaran, \ I \ get \ the \ impression \that \ you \ really \ want \ to \ learn \ "The \ Calculus",\)

so I am going to show you a few tips your teacher may have or may have not\displaystyle so \ I \ am \ going \ to \ show \ you \ a \ few \ tips \ your \ teacher \ may \ have \ or \ may \ have \ not

shown you. First we want to find the anti−dervative of ∫(x+5) (x−10)dx, don′t forget\displaystyle shown \ you. \ First \ we \ want \ to \ find \ the \ anti-dervative \ of \ \int\sqrt{(x+5)} \ (x-10)dx, \ don't \ forget

the dx.\displaystyle the \ dx.

Now, a cute trick. Whenever you have a radical with index 2 (in other words the square root\displaystyle Now, \ a \ cute \ trick. \ Whenever \ you \ have \ a \ radical \ with \ index \ 2 \ (in \ other \ words \ the \ square \ root

of something), let u = (x+5) = (x+5)1/2.\displaystyle of \ something), \ let \ u \ = \ \sqrt{(x+5)} \ = \ (x+5)^{1/2}.

This implies that du = 12(x+5)−1/2(1)dx = dx2(x+5)1/2.\displaystyle This \ implies \ that \ du \ = \ \frac{1}{2}(x+5)^{-1/2}(1)dx \ = \ \frac{dx}{2(x+5)^{1/2}}.

However, (x+5)1/2 = u, so we have du = dx2u or 2udu = dx, (we eliminated the\displaystyle However, \ (x+5)^{1/2} \ = \ u, \ so \ we \ have \ du \ = \ \frac{dx}{2u} \ or \ 2udu \ = \ dx, \ (we \ eliminated \ the

pesky radical).\displaystyle pesky \ radical).

Now, also note that u = (x+5)1/2   ⟹   that x = u2−5.\displaystyle Now, \ also \ note \ that \ u \ = \ (x+5)^{1/2} \ \implies \ that \ x \ = \ u^2-5.

Now, putting it all together, we get:\displaystyle Now, \ putting \ it \ all \ together, \ we \ get:

∫u(u2−5−10)2udu = 2∫u2(u2−15)du = 2∫(u4−15u2)du = 2[u55−5u3]+C.\displaystyle \int u(u^2-5-10)2udu \ = \ 2\int u^2(u^2-15)du \ = \ 2\int(u^4-15u^2)du \ = \ 2\bigg[\frac{u^5}{5}-5u^3\bigg]+C.

Now, resubbing, we get 2[(x+5)5/25−5(x+5)3/2]+C.\displaystyle Now, \ resubbing, \ we \ get \ 2\bigg[\frac{(x+5)^{5/2}}{5}-5(x+5)^{3/2}\bigg]+C.

This gives 2[(x+5)5/2−25(x+5)3/2]5+C = 2[(x+5)3/2((x+5)−25)]5+C.\displaystyle This \ gives \ \frac{2[(x+5)^{5/2}-25(x+5)^{3/2}]}{5}+C \ = \ \frac{2[(x+5)^{3/2}((x+5)-25)]}{5}+C.

Therefore, after all is said and done, we get: ∫(x+5)(x−10)dx = 2[(x+5)3/2(x−20)]5+C.\displaystyle Therefore, \ after \ all \ is \ said \ and \ done, \ we \ get: \ \int\sqrt{(x+5)}(x-10)dx \ = \ \frac{2[(x+5)^{3/2}(x-20)]}{5}+C.

Now, to elucidate furthur, how do we know we didn′t mess up somewhere along the line?\displaystyle Now, \ to \ elucidate \ furthur, \ how \ do \ we \ know \ we \ didn't \ mess \ up \ somewhere \ along \ the \ line?

Take a check, to wit: Dx[2[(x+5)3/2(x−20)]5+C] = x+5(x−10), QED.\displaystyle Take \ a \ check, \ to \ wit: \ D_x\bigg[\frac{2[(x+5)^{3/2}(x-20)]}{5}+C\bigg] \ = \ \sqrt{x+5} (x-10), \ QED.
 
Re: Integral of 3x/(Sqrt[4x+20])?

Thank you BigGlenn,

But if I started out with a C, I should end up with a Cx term, right?
 
Re: Integral of 3x/(Sqrt[4x+20])?

Jaskaran said:
\(\displaystyle \If\ I \ wanted \ to \ take \ the \ integral \ of \int\sqrt{x+5}\\(x-10)\+\C\ one \ more \ time, \ what\ would \ it \ look \ like?\\)

? ? (x+5)*(x-10) , u = x-10, du=dx

?? u+C

=

I get 2(x-10)^(1/2) + Cx
Click to expand...

u = x+5

∫u12(u−15)du = ∫u32du − 15∗∫u12du = 25u52 −  10∗u32 + C\displaystyle \int u^{\frac{1}{2}}(u-15)du \ = \ \int u^{\frac{3}{2}}du \ - \ 15*\int u^{\frac{1}{2}}du \ = \ \frac{2}{5}u^\frac{5}{2} \ - \ \ 10* u^\frac{3}{2} \ + \ C

and sub back.....
 
Re: Integral of 3x/(Sqrt[4x+20])?

Jaskaran, I′m losing patience with you.\displaystyle Jaskaran, \ I'm \ losing \ patience \ with \ you.

Look at Subhotosh Khan work, it is probably easier to understand.\displaystyle Look \ at \ Subhotosh \ Khan \ work, \ it \ is \ probably \ easier \ to \ understand.

But if I started out with a C, what is that.\displaystyle But \ if \ I \ started \ out \ with \ a \ C, \ what \ is \ that.
 
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