<p>How do you solve this using integration by parts? I have a feeling that it's really easy, but haven't gotten it. [Calc BC]</p>
<p>[integral]
<em>xe^(2x)</em> dx
(2x+1)^2</p>
<p>solution:
e^(2x)_ + C
4(2x+1)</p>
<p>Help appreciated =).</p>
<p>Cheers!</p>
<p>Seriously - no one?</p>
<p>You're uber genious if you get it ;)</p>
<p>use the tabular method.. mwahahhaa</p>
<p>Haha.. okay, tabular method. I'll accept - that's one way =P.</p>
<p>Now.. how do we do this algebraically? Hehe.</p>
<p>Thank for your input, beramod. Haha =P.</p>
<p>Well,I tried about three times and I got it. :)
Let * be the integral sign...</p>
<ul>
<li>xe^2x/(2x+1)^2 .dx</li>
</ul>
<p>Here's the imp. part..the substitution...</p>
<p>Let t=(2x+1) , Therefore, t^2=(2x+1)^2</p>
<p>t^2=4x^2+1+4x</p>
<p>2t.dt=8x+4.dx</p>
<p>2t.dt=4(2x+1),dx</p>
<p>t.dt=2(2x+1).dx, But (2x+1)=t</p>
<p>t.dt=2t.dx</p>
<p>Therefore, dx=dt/2</p>
<p>{ Keep in mind, x= (t-1)/2 }</p>
<p>Substituing the value of dx and x in the original eqn...</p>
<ul>
<li>[{((t-1)/2) . e^t-1}/2t^2].dt</li>
</ul>
<p>1/4 * (t-1. e^t-1) / t^2 .dt</p>
<p>1/4 * (1/t - 1/t^2) .e^t-1 .dt</p>
<p>1/4 * (1/t) . e^t-1 .dt - 1/4 * (1/t^2) . e^t-1 .dt</p>
<p>Integrating the first integral using By parts (Note the second integral will cut with the integral part of the by parts answer)</p>
<p>[ (1/4t). e^t-1 - 1/4 * -(1/t^2) . e^t-1 .dt ] - 1/4 * 1(/t^2) . e^t-1.dt</p>
<p>= (1/4t). e^t-1 </p>
<p>But t is 2x+1</p>
<p>= (1/4(2x+1)) . e^2x //</p>
<p>That's the answer...note ...there is no C (constant) in the answer as we don't solve any integrals..</p>
<p>Hope what I wrote makes sense.....</p>
<p>Roh kay.</p>
<p>Integral xe^2x/(2x+1)^2 dx by parts.</p>
<p>integral (udv) = uv - integral vdu</p>
<p>So we have [Int.] xe^2x*(2x+1)^-2 dx</p>
<p>u = xe^2x
dv = (2x+1)^-2 dx</p>
<p>Solving by simple substitution we can find du and v:</p>
<p>du = xdx*e^2x + e^2x dx * x
du = e^2x + 2xe^2x = (2x+1)e^2x</p>
<p>v = integral of (2x+1)^-2 dx</p>
<p>Substitution time. t = 2x+1, dt=2dx, dt/2=dx</p>
<p>Integral (t^-2/2 dt) = 1/2 [Integral] t^-2 = 1/-2t</p>
<p>Substitute back:
v = 1/-2(2x+1)</p>
<p>Now we know all of our parts. I'm simplifying a little bit and plugging back into the int.udv=uv-int.vdu formula</p>
<p>Integral xe^2x(2x+1)^-2 = xe^2x/-2(2x+1) - <a href="2x+1">integral</a>e^2x/-2(2x+1)</p>
<p>The 2x+1 cancel in the integral leaving:</p>
<p>xe^2x/-2(2x+1) - [Integral] e^2x/-2 =</p>
<p>""""""""""""""""""""" + 1/2[Integral] e^2x (Use subst. for u and say u=2x)</p>
<p>du/2=dx, [Integral] e^u/2 = 1/2[Integral] e^u = e^u/2 = e^2x/2</p>
<p>So we have:</p>
<p>xe^2x/-2(2x+1) + 1/2(e^2x/2) =</p>
<p>""""""""""""""""""""" + e^2x/4 =</p>
<p>-2xe^2x/4(2x+1) + (2x+1)e^2x/4(2x+1) =</p>
<p>(-2xe^2x + 2xe^2x + e^2x)/4(2x+1) =</p>
<p>e^2x/4(2x+1) + C</p>
<p>I love you all.</p>
<p>Geez I almost amazed myself with that beautiful art of mathematics!</p>
<p>atreeyum- come look and see what i did!!!! lol</p>
<p><em>nt</em> : I was headed in that direction but messed up somewhere because the substitution didn't work for some reason, hehe. It all got tangled =X... Thanks for your time!</p>
<p>adidasty: I agree, math is beautiful, haha. Isn't it a great feeling to spend some time on a perplexing problem and get the write answer... or just understand it completely? Haha. Wow, nice work =). Thanks for your help!</p>
<p><em>nt</em> and adidiasty = uber genius =P, haha.</p>
<p>Cheers! Ya'll are awesome =D!</p>
<p>That's the answer...note ...there is no C (constant) in the answer as we don't solve any integrals..</p>
<p>What are you talking about?</p>
<p>a particle is moving to the right.
x(t)=cos(pi*t^2) 0 less than or equal to t less than or equal to 3 </p>
<p>what is the velocity as a function of t?
what is the acceleration as a function of t?
what are the values when it is moving right?
what is the acceleration at an instant when it returns to t?</p>
<p>x(t) = cos(pi*t^2)</p>
<p>Velocity = v(t) = -2pi<em>t[sin(pi</em>t^2)]</p>
<p>Acceleration = a(t) = -2pi[sin(pi<em>t^2)] - 2pi</em>t[cos(pi*t^2)]</p>
<p>Find values where a(t) is greater than 0 and it is moving to the right.</p>
<p>I dunno the last one.</p>
<p>Azmodan...well......when i opened the integral by parts,both the integrals cancelled...............so I didn't solve any integral sign. So no C.</p>
<p>I have no idea what you did, but regardless, you can't integrate without having a C at the end, unless you do it wrong.</p>
<p>Math is the most beautiful art ever</p>
<p>hm..Azmodan...Please check my calculations. I have integration this year,and it's for about 30 marks in my paper..so I'd better get to know my mistakes :P</p>
<p>I think something might be wrong...or maybe in this case C=0.</p>
<p>I'm not sure what you did, the *'s make it too confusing.</p>
<p>Anyway, all integrating is is anti-differentiating. Therefore, every indefinate integral should end with a + C because the constant will drop out if it's differentiated.</p>