Ok, I know you do this by parts but i don't see how... how do you integrate (4x)(e^[x squared]) and apologies, i still have no idea how to write this nicely, in a math algebra-ish way
Integration by parts relies on the product rule of derivatives: \(D[u v] = u D[v] + D[u]v = u v' + u' v\) leads to \(uv =\int u v' dx \, + \, \int u' v dx\) or \(\int u v' dx = uv \, - \, \int u' v dx\) If you were to try to solve it by integration by parts you would write (following the LIPET heuristic) \(u = 4x, v' = e^{x^2}\) giving \(u' = 4\) and a complicated odd function for v. \(v(x) = \frac{\sqrt{\pi}}{2} \textrm{erfi}(x)\). Thus we get \(\int \, 4 x e^{x^2} \, dx = 2 \sqrt{\pi} x \textrm{erfi}(x) - \int \, 2 \sqrt{\pi} \textrm{erfi}(x) \, dx = 2 \sqrt{\pi} x \textrm{erfi}(x) - 2 \sqrt{\pi} x \textrm{erfi}(x) + 2 e^{x^2} + C = 2 e^{x^2} + C\) which is horrible if you aren't familiar both with erfi() and its integral. http://mathworld.wolfram.com/Erfi.html If you knew the definition \(e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}\) you could rely on the linearity of integration which follows from the linearity of derivatives: \(D[a u + b v] = au' + bv'\). Thus \(\int \, 4 x e^{x^2} \, dx = \sum_{k=0}^{\infty} \int \, 4 \frac{x^{2k+1}}{k!} \, dx = C + \sum_{k=0}^{\infty} 4 \frac{x^{2k+2}}{2 (k+1) \cdot k!} = C_1 + \sum_{k=0}^{\infty} 2 \frac{x^{2(k+1)}}{(k+1)!} = C_1 + 2 \sum_{k=1}^{\infty} \frac{(x^2)^k}{k!} = C_1 + 2 \left( e^{x^2} - 1 \right) = e^{x^2} + C\) But as Tach correctly points out, the simplest way to solve this problem is via u-substitution which follows from the chain rule of derivatives. If \((v \circ u)\) is the composition of functions such that \((v \circ u)(x) = v(u(x))\) then the chain rule is \(D[v \circ u ] = ( v' \circ u ) u'\). So \(D[v(u(x))] = v'(u(x)) \cdot u'(x)\) and if we find a integral that looks like \(\int \, v'(u(x)) \cdot u'(x) \, dx\) we know the answer is \(v(u(x)) + C\). Here \(u(x) = x^2, \, u'(x) = 2x \) and by the process of elimination, \(v'(x) = 2 e^x = v(x)\) so \(\int \, 4 x e^{x^2} \, dx = \int \, 2 e^{x^2} \cdot (2x) \, dx = \int \, 2 e^{u(x)} u'(x) \, dx = 2 e^{u(x)} + C = 2 e^{x^2} + C\)
Sometimes u-substitution can be tricky to find. \(\int \, \frac{x^2 -1 }{\left( x^2 + 1 \right)^2} \, dx\) \(\int \, \tan x \, dx\) \(\int \, \frac{33 x^5 + 24}{3 x^6 + 4x} \, dx\)
Finding the correct approach is more like black magic, isn't it? This makes integration and solving differential equations so much fun.
Yes it is like black magic! I still have no idea how you did that with the substitution, but I found an even better way after staring at it for hours! It turns out if you take the 2 outside, then \( 2x \) becomes part of the integral of \( e^ {x^2} \) which simplifies nicely. Aaaah maths! Please Register or Log in to view the hidden image! and thank you tach!!! and rpenner! Please Register or Log in to view the hidden image!