# Find the Indefinite integral using integration by parts: S(x^2)(e^x) dx using the equation Sudv = uv - Svdu I have  figured out that u=x^2 and dv= e^x dx and du = dx and v=e^x.  Assuming that is right, I have calculated the following: (x^2)(e^x) - Se^x dx (x^2)(e^x) - e^x +C (e^x)((x^2) -1) +C

Luca B. | Certified Educator

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You need to use the following formula to integrate by parts such that:

int udv = uv - int vdu

You have made a right selection considering u = x^2 , since differentiating it, its degree will minimize, but you did not differentiate it right. You should follow the next formula when differentiate x^2  such that:

(x^n)' = n*x^(n-1)

Reasoning by analogy yields:

(x^2)' = 2*x^(2-1) = 2x

Hence, considering u = x^2  and dv = e^x dx  yieldS:

u = x^2 => du = 2xdx

dv = e^x dx => v = e^x

int x^2 e^x dx = x^2*e^x - int 2x*e^x dx

int x^2 e^x dx = x^2*e^x - 2 int x* e^x dx

You need to use parts againto sol,ve the integral int x*e^x dx  such that:

u = x => du = dx

dv = e^xdx => v = e^x

int x* e^x dx = x*e^x - int e^x dx

int x* e^x dx = x*e^x - e^x + c

Substituting x*e^x - e^x  for int x* e^x dx  yields:

int x^2 e^x dx = x^2*e^x - 2 (x*e^x - e^x) + c

int x^2 e^x dx = x^2*e^x - 2e^x*(x - 1) +  c

Factoring out e^x  yields:

int x^2 e^x dx = e^x*(x^2 - 2x + 2) + c

Hence, evaluating the given indefinite integral using parts yields int x^2 e^x dx = e^x*(x^2 - 2x + 2) + c .

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widowspeak | Student

You are correct to take u=x^2 and dv = e^x. However, du = 2x dx (i.e. the derivative of x^2 is 2x not 1).

So you should get

x^2 e^x - S 2x e^x dx

S 2x e^x dx itself needs to be integrated by parts

Now u=2x dv=e^x, meaning du = 2, v = e^x

giving S 2x e^x dx = 2x e^x - 2 S e^x dx = 2x e^x - 2e^x + C

Putting this result into the first equation gives

x^2 e^x - 2x e^x + 2e^x + C