# How to find the integral of x 3^(x^2)

## Homework Statement

$$\int$$ x 3^(x^2)

## Homework Equations

integration by parts

## The Attempt at a Solution

u= x dv= 3^(x^2)
du = 1 dx v=((3^(x^2))/ln 3 ) 2x + c ??

(x)((3^(x^2))/ln 3 ) 2x - $$\int$$ ((3^(x^2))/ln 3 ) 2x dx

is this right??

## Homework Statement

$$\int$$ x 3^(x^2)

## Homework Equations

integration by parts

## The Attempt at a Solution

u= x dv= 3^(x^2)
du = 1 dx v=((3^(x^2))/ln 3 ) 2x + c ??
No: what do you get when you differentiate your v? 3^(x^2) doesn't have a nice integral.

Rather, you should recognise $$\int x3^{x^2} \, dx$$ is built for a substitution: let $$u = x^2$$.

so u=x^2
du = 2x

1/2 3^u du??

Hitman2-2

so u=x^2
du = 2x

1/2 3^u du??

do you mean

$$\frac{1}{2}\int 3^u du$$

Hitman2-2

Unco is correct: use the substitution

HallsofIvy
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