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How do you integrate #intx\(3^(x^2+1))dx#?

Answer 1

Isabella Ackerman

The answer is #=1/2*3^(x^2+1)/ln3 +C#

Let #u=x^2+1#, #=>#, #du=2xdx#

Therefore,

#I=intx(3^(x^2+1))dx=1/2int3^udu#

Let #v=3^u#

#lnv=ln(3^u) = u ln3#

#v=e ^(u ln3)#

Therefore,

#I=1/2inte ^(u ln3)du #

#=1/2*e ^(u ln3)/(ln3)#

#=1/2*3^(x^2+1)/ln3 +C#

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Answer 2

Elijah Abram

To integrate (\int 3^{x^2+1} dx), you can use a substitution method. Let (u = x^2 + 1). Then (du = 2x dx). Solving for (dx), we get (dx = \frac{1}{2x} du).

Substitute (u = x^2 + 1) and (dx = \frac{1}{2x} du) into the integral:

(\int 3^{x^2+1} dx = \int 3^u \cdot \frac{1}{2x} du)

Now, since (u = x^2 + 1), we need to express (x) in terms of (u). (x^2 = u - 1), so (x = \sqrt{u - 1}).

Substitute (x = \sqrt{u - 1}) into the integral:

(\int 3^u \cdot \frac{1}{2\sqrt{u-1}} du)

Now, this integral can be solved using standard techniques.

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