How do you integrate #(e^x/x)dx - Socratic
WEBmason m. Apr 12, 2017. This is sometimes called the exponential integral: ∫ ex x dx = Ei(x) +C. But the method I'd use (since I'm not familiar with the integral) is the Maclaurin series for ex: ex = 1 + x + x2 2! + x3 3! +... = ∞ ∑ n=0 xn n! Then: ex x = 1 x + 1 + x 2! + x2 3! +... = 1 x + ∞ ∑ n=0 xn (n + 1)! So the antiderivative ...
How do you find the integral of e^(-x)dx? - Socratic
WEBMay 15, 2018 · Explanation: let u = −x then du dx = − 1 which means dx = − du. then. ∫(e−x)dx = ∫(− eu)du = −∫(eu)du = − eu +c = −e−x +c. Answer link.
Integrals of Exponential Functions - Calculus - Socratic
WEBThe answer is #I=-e^ (-x)+C#. This integral can be solved by a substitution: #u=-x# #du=-dx# #-du=dx#. So, now we can substitute: #int e^ (-x)dx = int e^u (-du)# #=-int e^u du# #=-e^u + C#. and substitute back: #=-e^ (-x) + C#. For simple looking integrands, you should try a quick check to see if substitution works before trying harder ...
How do you find the integral of #e^(x^2)#? - Socratic
WEBApr 27, 2015 · Since ex = 1 + x + x2 2! + x3 3! +⋯ = 1 + x + x2 2 + x3 6 +⋯ (for all x), it follows that ex2 = 1 +x2 + x4 2 + x6 6 + ⋯ (for all x). It is valid in this example to now integrate term-by-term (the result is true for all x): ∫ex2dx = ∫(1 + x2 + x4 2 + x6 6 +⋯)dx. = C + x + x3 3 + x5 10 + x7 42 +⋯. Alternatively, you can also give ...
How do you integrate (x^2)(e^x)dx? - Socratic
WEBSep 17, 2016 · Putting this in (1), we get. ∫x2exdx = x2ex −2(xex −ex) +c. = ex(x2 −2x + 2) + c. intx^2e^xdx=e^x (x^2-2x+2)+c We do it using integration by parts. Let u=x^2 and v=e^x, then du=2xdx and dv=e^xdx Now integration by parts states that intu (x)v' (x)dx=u (x)v (x)-intv (x)u' (x)dx Hence intx^2e^xdx=x^2e^x-inte^x xx 2xdx = x^2e^x-2intxe^xdx+c
How do you find the integral of (e^x)(cosx) dx? - Socratic
WEBJun 22, 2015 · In terms of complex numbers, ex is just some real factor, so it doesn't matter whether we have it outside or inside the Real part function. This means we can put the entire integral inside the Real part function: = Re(∫ exeix dx) = Re(∫ ex+ix dx) = Re(∫ e(i+1)x) dx =. We can do a quite simple u-substitution to evaluate the integral:
How do you integrate # (1/(e^x+1))dx - Socratic
WEBMar 25, 2017 · So the integral can be expressed as: ∫ 1 ex + 1 dx = x −ln(ex + 1) + C. x-ln (e^x+1)+C Let e^ (x/2)=tantheta. Then 1/2e^ (x/2)dx=sec^2thetad theta. intdx/ (e^x+1)=2int (1/2e^ (x/2)dx)/ (e^ (x/2) (e^x+1))=2int (sec^2thetad theta)/ (tantheta (sec^2theta))=2intcostheta/sinthetad theta =2lnabssintheta From tantheta=e^ (x/2) draw …
How do you integrate int x^3 e^x dx using integration by parts?
WEBMar 1, 2016 · Using the common integral ∫exdx = ex. = 3(x2ex − 2(exx −ex)) = x3ex −3(x2ex −2(exx − ex)) Adding a constant to the solution, = x3ex −3(x2ex −2(exx − ex)) +C. Answer link. \int x^3e^xdx=x^3e^x-3 (x^2e^x-2 (e^x x-e^x))+C \int x^3e^xdx Applying integration by parts as: \int uv'=uv-\int u'v u=x^3,u'=3x^2,v'=e^x,v=e^x =x^3e^x-\int ...
Integration by Parts - Calculus - Socratic
WEBint x*e^x dx is the integral we need to evaluate. If we use e^x as the first function and x as the second and integrate by parts, int x*e^x*dx = e^x int x*dx - int (e^x int x*dx)*dx. = e^x*x^2/2 - int e^x*x^2/2*dx + C. If we apply integration by parts to the second term, we again get a term with a x^3 and so on.
How do you integrate #int x^2 e^(x^2 ) dx # using ... - Socratic
WEBApr 30, 2016 · See the explanation section below. To integrate x to a power times e to a power, we expect to differentiate the x and integrate the e to a power int x^2 e^ (x^2 ) dx In order to integrate e^ (x^2) dx we need an x so that we can use substitution. int x^2 e^ (x^2) dx = int x e^ (x^2)x dx . Let u = x and dv = e^ (x^2)x dx The du = 1 dx and v = 1/2 ...