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How do you integrate #int 3xe^(x^2+1) dx#?

Answer 1

In this case, the integrand is the derivative of a function composition (up to a multiplicative constant): #g(y)=e^y# #f(x)=x^2+1# #rArr g(f(x))=e^(x^2+1)# #rArr d/(dx) [g(f(x))]=d/(dx)[e^(x^2+1)]=2x e^(x^2+1)#

The only difference is the multiplicative constant (it's #2# instead of #3#), so we can rewrite the integral to look like the derivative of #e^(x^2+1)# by working with multiplicative constants: #int 3x e^(x^2+1) dx=3/2 int 2x e^(x^2+1) dx=3/2 e^(x^2+1) + C#

Another (very similar) way of solving this was to consider the substitution #u(x)=x^2+1#. We could relate the two differentials #du=2x dx# and rewrite the integral in terms of #u#: #int 3x e^(x^2+1) dx=int 3/2 e^(x^2+1) (2xdx)=int 3/2 e^u du=3/2e^u+C# Now we could return back to #x#: #3/2e^u+C=3/2 e^(x^2+1)+C#

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Answer 2

Emilia Aguilar

To integrate ∫3xe^(x^2+1) dx, use u-substitution. Let u = x^2 + 1, then du = 2x dx. This gives us 3/2 times the integral of e^u du. Integrating e^u gives us e^u, so the final result is 3/2 e^(x^2+1) + C, where C is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.