I dont know the code for exponential

[frozen7]

\int \ell^{x^2} x^2 dx

= 1/2 \int \ell^{x^2} x^2 d(x^2)
= 1/2 \int \ell^{t} . t dt (Let x^2 = t )
= 1/2 [t.\ell^{t} - \int \ell^{t} dt]
= 1/2 [t.\ell^{t} - \ell^{t} + c ]
= 1/2 [x^2 \ell^{x^2} - \ell^{x^2} + c]

\ell is actually exponential. I don't know the code for exponential. :)
What`s the problem with the above integration I have done?

 

[d_leet]

\int \ell^{x^2} x^2 dx
= 1/2 \int \ell^{x^2} x^2 d(x^2)
= 1/2 \int \ell^{t} . t dt (Let x^2 = t )
= 1/2 [t.\ell^{t} - \int \ell^{t} dt]
= 1/2 [t.\ell^{t} - \ell^{t} + c ]
= 1/2 [x^2 \ell^{x^2} - \ell^{x^2} + c]
\ell is actually exponential. I don't know the code for exponential. :)
What`s the problem with the above integration I have done?

Well you set t= x^2 but you don't have d(x^2) in the original integral, just integrate by parts you don't really have to make a substitution.

 

[VietDao29]

\int \ell^{x^2} x^2 dx
= 1/2 \int \ell^{x^2} x^2 d(x^2)
= 1/2 \int \ell^{t} . t dt (Let x^2 = t )
= 1/2 [t.\ell^{t} - \int \ell^{t} dt]
= 1/2 [t.\ell^{t} - \ell^{t} + c ]
= 1/2 [x^2 \ell^{x^2} - \ell^{x^2} + c]
\ell is actually exponential. I don't know the code for exponential. :)
What`s the problem with the above integration I have done?

The error lies in line #2, you are changing dx into d(x²), that requires an x, that'll left you with:
\frac{1}{2} \int x e ^ {x ^ 2} d(x ^ 2) not \frac{1}{2} \int x ^ 2 e ^ {x ^ 2} d(x ^ 2). Or you can do a little bit more slowly by letting t = x². Just try it, and see if you can get to:
\frac{1}{2} \int x e ^ {x ^ 2} d(x ^ 2).
From here, you can do the same and end up with:
\frac{1}{2} \int x e ^ {x ^ 2} d(x ^ 2) = \frac{1}{2} \int x d(e ^ {x ^ 2}).
From here, do you know what to do next?