I don't understand how to solve this problem I tried using u-sub and integration by parts

[granite cipher]

lovely integral not gonna like

anyways, the secret sauce to solving this juicy integral boy is to realize that

$e^{2x} = (e^x)^2$

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tooBoard

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$u = e^x$

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$du = e^x dx$

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$\int (u^2 + 1)^3 dx$

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do you want me to continue @zealous marlin

The rest should be familiar, , honestly this was a fun integral to solve

[zealous marlin]

@granite cipher please do

I don't know how to take the antiderivative of something like this

[subtle wharf]

It should be du but prob a typo

[subtle wharf]

you can use his substution and then expand the ^3 that is managble you will get polynomial

[zealous marlin]

I think I did it

I ended up wih (e^2x+1)^4/4e^2x +c but I don't know if it's right

[subtle wharf]

how did you expand (u²+1)³

[zealous marlin]

mental math tbh

just reverse chain ruling

[granite cipher]

so be it

$\int (u^2 + 1)(u^2 + 1)^2dx$

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$\int (u^2 + 1)(u^4 + 2u^2 + 1)dx$

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$\int (u^6 + 2u^4 + u^2 + u^4 + 2u^2 + 1)$

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$\int (u^6 + 3u^4 + 3u^2 + 1)$

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$\frac{1}{7}u^7 + \frac{3}{5}u^5 + u^3 + u + C$

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remember $u = e^x$

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tooBoard

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$\therefore$

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$\frac{1}{7}e^{7x} + \frac{3}{5}e^{5x} + e^{3x} + e^x + C$

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tooBoard

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$\int e^x(e^{2x} + 1)^3 dx = \frac{1}{7}e^{7x} + \frac{3}{5}e^{5x} + e^{3x} + e^x + C$

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tooBoard

[granite cipher]

@zealous marlin