Evaluate the integrals in Exercises 31–56. Some integrals do n...

Question

Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.

∫ (xe^x) / (x + 1)² dx

Accepted Answer

Welcome back everyone. Evaluate the integral x minus 1 e to the power of x d x divided by x 2. For this problem, let's begin by distributing. So. If we distribute, we're going to get integral of x e to the power of x divided by x2 minus e to the power of x divided by x2 followed by dx. Now notice that we can simplify x divided by x 2 gives us x in the denominator, so we get integral of e to the power of x divided by x minus e to the power of x divided by x 2. And now we can apply the properties of integrals to rewrite it as a difference of two integrals, integral of e to the power of x divided by x dx minus integral of e to the power of x divided by x 2 dx. Let's begin with the first integral and let's use integration by parts. Remember that the integral of UDV is equal to UV minus integral of VDU. And we can basically set u equals 1 divided by x. We can set dV is equal to e to the power of x dx, which is the remaining part of the integral, and now let's identify duu. By taking the derivative of u with respect to x, so DU divided by dx is the derivative of 1 divided by X, which is -1 divided by x 2. And now we can solve for DU, right? We can show that DU is equal to negative DX divided by X2 if we multiply both sides by DX. So now that we have d u, let's identify v, which is the integral of dv, and as the integral of eats the power of x, d x, which is e to the power of x. No constant of integration is required for now. We can just go ahead and apply the formula. So we get integral of e to the power of x divided by x d x equals u multiplied by v. So that'd be 1 divided by X. Multiplied by each to the power of x. Minus integral of VDU. So that would be Minus We can add negative. DX divided by X 2 multiplied by e to the power of x, right? Well done, and from here we can factor out the constant. Which gives us e to the power of x divided by x for the first term minus negative gives us positive sign, right? Integral of e to the power of x divided by x 2 dx. Well done, and now we can substitute this result into the. Integrals, specifically the original integral where we have a difference of two separate integrals. We can now show that the original integral. E to the power of x divided by x dx minus integral of e to the power of x divided by x 2 dx is equal to. The first integral is going to be e to the power of x divided by x plus integral of e to the power of x divided by x 2 dx. Minus. The remaining integral is is the power of x. Divided by x 2 d x. Notice that we're going to cancel out the two integrals because we have opposite signs and we basically get e to the power of x divided by x and we're going to add a constant of integration c, right? Because we have Indefinite integrals. Well done, so we have our final answer. It's the power of x divided by x plus c. Let's write it in a clear form. It's the power of x divided by x plus c. That's our final answer, and thank you for watching.

Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.∫ (xe^x) / (x + 1)² dx

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Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.
∫ (xe^x) / (x + 1)² dx