What is the integration of e^√x?
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What is the integration of e^x*(x-1) /x^2?
Let [math]I=\int e^x\dfrac{(x-1)}{x^2}dx[/math][math]=\int e^x[\dfrac{1}{x}-\dfrac{1}{x^2}]dx[/math][math]=\int\dfrac{e^x}{x}dx-\int\dfrac{e^x}{x^2}dx[/math][math]=\int\dfrac{e^x}{x}dx+\dfrac{e^x}{x}-\int\dfrac{e^x}{x}dx+C[/math][math]=\dfrac{e^x}{x}+C[/math]Where 'C' being integrating constants.
What is the integration of e^ (-x^2) dx over the limit from 0 to 1? How do you solve it?
Put x^2=t, then dx=1/2dtNow change the limits by substituting in x^2=tLimits are 0 to 1 same as previous limitsNow integrate,By substituting in integral we get,Integral e^[-t] dt=-e^(-t) limits from 0 to 1By substituting we get, - (e^[-1] -1)=1-1/e=(e-1)/e
What's the integration of e^ (x^3/3) /x^2?
Say, [math]I = \int e^{\frac{x^3}{3}} \cdot x^{2} [/math][math]\, dx[/math]Let [math]t = \frac{x^3}{3} \implies dt = x^2 \, dx[/math]So, [math]I = \int e^t \, dt = e^t + c = e^{\frac{x^3}{3}} + c[/math]
How can I solve the integration of, e^x (1+ sin x / 1+ cos x) dx?
[math]\large\displaystyle\star[/math] A2AEvaluate:[math]\large\displaystyle I = \large\displaystyle \int e^x \left(\frac{1 + \sin x}{1 + \cos x}\right) \, dx[/math][math]\implies\large\displaystyle I = \large\displaystyle\int e^x \left(\frac{1 + 2 \sin \left(\frac{x}{2} \right) \cos \left(\frac{x}{2}\right)}{2 \cos^2 \left(\frac{x}{2} \right)}\right) \, dx[/math][math]\implies\large\displaystyle I = \large\displaystyle\int e^x \left(\frac{1}{2} \sec^2 \left(\frac{x}{2} \right) + \tan \left(\frac{x}{2}\right)\right) \, dx[/math][math]\large\displaystyle\star[/math] We know that:[math]\large\displaystyle\int e^x [f(x) + f'(x)] \, dx = \large\displaystyle e^x f(x) + C[/math][math]\implies\large\displaystyle I = \large\displaystyle \int e^x \left(\large\displaystyle\underbrace{\tan \left(\frac{x}{2}\right)}_{\large\displaystyle f(x)} + \large\displaystyle\underbrace{\frac{1}{2} \sec^2 \left(\frac{x}{2} \right)}_{\large\displaystyle f'(x)} \right) \, dx[/math][math]\implies\boxed{\boxed{\large\displaystyle I = \large\displaystyle e^x \tan \left(\frac{x}{2}\right) + C}}[/math]Thanks![math]\large\displaystyle\boxed{\huge{\huge{\displaystyle\ddot\smile}}}[/math]
What is the integral of e^x/e^2x-3e^x+2?
Sub u=e^x and du=e^x dx so integral [e^x/(e^2x-3e^x+2)] dx =integral [1/(u^2-3u+2)]du and using partial fractions 1/(u-2) -1/(u-1) du you get ln(u-2)-ln(u-1)=ln(e^x-2)-ln(e^x-1)+C
Integral of e^(-x^2)?
It's the error function (also called the Gauss error function) It is defined as: erf(x) = (2/√π)∫ (from 0 to x) e^(-t^2) dt ----> ∫ (from 0 to x) e^(-t^2) dt =[ (√π)erf(x)]/2 http://mathworld.wolfram.com/Erf.html
What is the integral of [math]e^{2x}[/math]?
The integral of [math]e^u du[/math] is [math]e^u + c[/math].Looking at the expression above, we can see that [math]u = 2x, du = 2dx[/math]. But we do not have a 2 in our expression. Let's add it.[math]\int{e^{2x}}dx =[/math][math]\frac{1}{2}\int{e^{2x}2dx} = [/math][math]\frac{1}{2}e^u + c = [/math][math]\frac{1}{2}e^{2x} + c[/math]So there we are.
What is the integral of e^x (2+sin 2x) dx/cos^2(x)?
Given :-[math]\qquad \displaystyle\int \dfrac{e^x(2+sin2x)}{cos^2x} dx[/math][math]\qquad \displaystyle\int \dfrac{e^x(2+2sinxcosx)}{cos^2x} dx[/math]Take [math]2[/math] common and separate the denominator, u will get :-[math]\qquad 2\displaystyle\int e^x(sec^2x + tanx) dx[/math]The above form will match with :-[math]\qquad \displaystyle\int e^x(f(x) + f'(x)) dx = e^xf(x) + c[/math]So we will get :-[math]\qquad 2e^xtanx + C [/math]