Integration by parts practice problems

1. 1
   Choose u and dv

   Differentiate the polynomial and integrate the exponential.

   $$u=x,\ du=dx,\ dv=e^x dx,\ v=e^x$$
2. 2
   Apply the formula

   Use uv minus the integral of v du.

   $$xe^x-\int e^x dx=e^x(x-1)+C$$

Independent check: The product rule differentiates eˣ(x - 1) to xeˣ.

1. 1
   Set the parts

   Let the polynomial be u.

   $$u=x,\quad v=-\cos x$$
2. 2
   Substitute

   Apply uv minus integral v du.

   $$-x\cos x+\int\cos x\,dx$$
3. 3
   Finish

   Integrate cosine.

   $$-x\cos x+\sin x+C$$

Independent check: Differentiation cancels the two cosine terms and leaves x sin x.

1. 1
   Expose a product

   Treat the integrand as ln x times 1.

   $$u=\ln x,\quad dv=dx$$
2. 2
   Find du and v

   Differentiate the logarithm and integrate 1.

   $$du=\frac1x dx,\quad v=x$$
3. 3
   Apply and simplify

   The new integral becomes the integral of 1.

   $$x\ln x-\int1\,dx=x\ln x-x+C$$

Independent check: The product rule gives ln x + 1 - 1 = ln x.

1. 1
   First application

   Differentiate x squared and integrate e to the x.

   $$x^2e^x-2\int xe^x dx$$
2. 2
   Evaluate the remaining integral

   A second application gives e to the x times x minus 1.

   $$\int xe^x dx=e^x(x-1)$$
3. 3
   Combine

   Factor e to the x after substitution.

   $$e^x(x^2-2x+2)+C$$

Independent check: Differentiating the product cancels the lower-degree terms and leaves x²eˣ.

1. 1
   Set the parts

   Differentiate x and integrate cosine of 2x.

   $$u=x,\quad v=\frac12\sin(2x)$$
2. 2
   Apply

   The remaining integral is one half the integral of sine of 2x.

   $$\frac{x}{2}\sin(2x)-\frac12\int\sin(2x)dx$$
3. 3
   Finish

   The sine antiderivative supplies another negative sign.

   $$\frac{x}{2}\sin(2x)+\frac14\cos(2x)+C$$

Independent check: Differentiation cancels the sine terms and returns x cos(2x).

1. 1
   Find the antiderivative

   Integration by parts gives the checked form.

   $$F(x)=e^x(x-1)$$
2. 2
   Evaluate the endpoints

   Keep the subtraction of the lower endpoint explicit.

   $$F(1)-F(0)=0-(-1)=1$$

Independent check: The integrand is nonnegative on [0, 1], and numerical quadrature gives 1 to the displayed precision.

1. 1
   Choose the parts

   Differentiate arctangent and integrate 1.

   $$u=\arctan x,\ du=\frac{dx}{1+x^2},\ v=x$$
2. 2
   Apply

   The new integral is handled by a simple substitution.

   $$x\arctan x-\int\frac{x}{1+x^2}dx$$
3. 3
   Finish

   Let w equal 1 plus x squared.

   $$x\arctan x-\frac12\ln(1+x^2)+C$$

Independent check: The product derivative and logarithm derivative cancel the extra x/(1 + x²) term.

1. 1
   Choose

   Differentiate the logarithm and integrate x.

   $$u=\ln x,\quad v=\frac{x^2}{2}$$
2. 2
   Apply

   The x in du cancels one power of x.

   $$\frac{x^2}{2}\ln x-\frac12\int x\,dx$$
3. 3
   Finish

   Integrate the remaining linear term.

   $$\frac{x^2}{2}\ln x-\frac{x^2}{4}+C$$

Independent check: Differentiation leaves x ln x after the extra x/2 terms cancel.

1. 1
   Apply parts once

   Integrate cosine and differentiate the exponential.

   $$I=e^x\sin x-\int e^x\sin x\,dx$$
2. 2
   Apply parts again

   The second integral produces negative e to the x cosine x plus I.

   $$\int e^x\sin x\,dx=-e^x\cos x+I$$
3. 3
   Solve for I

   Collect both copies of the original integral.

   $$2I=e^x(\sin x+\cos x)$$

Independent check: The product rule simplifies the derivative to eˣ cos x.

1. 1
   Set u and v

   The logarithm simplifies under differentiation.

   $$u=\ln x,\quad du=\frac1x dx,\quad v=\frac{x^3}{3}$$
2. 2
   Apply

   Cancel one x in the remaining integral.

   $$\frac{x^3}{3}\ln x-\frac13\int x^2dx$$
3. 3
   Finish

   Integrate x squared.

   $$\frac{x^3}{3}\ln x-\frac{x^3}{9}+C$$

Independent check: Differentiating produces x² ln x because the two x²/3 terms cancel.