Test 2 Review Material

Test Instructions

Below are the general instructions for all tests

  • You will get a chance to retake this. Your highest score counts toward your final grade.
  • Follow the guidance for each part and show all work for full credit.
  • Non-graphical calculators are allowed.
  • One page (two sides) of handwritten (or font size 8) notes are allowed.
  • The equation sheet from the textbook is included as part of the test pack (it doesn’t count as note pages)

Test Format

The Format of the test will include:

  • 10 pick the best technique (do not solve) [10pts]
  • 2 Substitution [2 x 5pts]
  • 2 Integration by Parts [2 x 5pts]
  • 1 Partial Fraction [5pts]
  • 3 Choose your method and solve [3 x 5pts]

Practice: Identifying the Appropriate Technique

For each integral below, choose the single most appropriate method.
You do not need to evaluate the integral.

Options:

  • A. Basic integration rules only
  • B. Substitution
  • C. Integration by parts
  • D. Partial fractions

  1. \[ \int (2x+3)^5\,dx \]

  2. \[ \int x e^{4x}\,dx \]

  3. \[ \int \frac{3}{x^2-16}\,dx \]

  4. \[ \int \ln(5x)\,dx \]

  5. \[ \int \frac{8x}{x^2+1}\,dx \]

  6. \[ \int (x^2-1)^3\,dx \]

  7. \[ \int \frac{2x-5}{x^2-5x+4}\,dx \]

  8. \[ \int x^3 \ln x\,dx \]

  9. \[ \int \frac{7x+1}{x(x+1)}\,dx \]

  10. \[ \int e^{-2x}\,dx \]

  11. \[ \int x\cosh x\,dx \]

  12. \[ \int \frac{5}{(x-1)(x+4)}\,dx \]

  13. \[ \int x^2 e^{x}\,dx \]

  14. \[ \int \frac{4x}{\sqrt{x^2+9}}\,dx \]

  15. \[ \int (3x^4-2x)^2\,dx \]

  16. \[ \int x\ln(x^2)\,dx \]

  17. \[ \int \frac{6x+3}{x^2+3x+7}\,dx \]

  18. \[ \int x^2\arctan x\,dx \]

  19. \[ \int \frac{2}{x^2-1}\,dx \]

  20. \[ \int x e^{-x}\,dx \]

Click to reveal answer key

Answer Key — Technique Only

  1. B (substitution)
  2. C (integration by parts)
  3. D (partial fractions)
  4. C (integration by parts)
  5. B (substitution)
  6. A (basic rules only)
  7. B (substitution — numerator is derivative of denominator)
  8. C (integration by parts)
  9. D (partial fractions)
  10. A (basic rules only)
  11. C (integration by parts)
  12. D (partial fractions)
  13. C (integration by parts)
  14. B (substitution)
  15. A (basic rules only)
  16. C (integration by parts)
  17. B (substitution)
  18. C (integration by parts)
  19. D (partial fractions)
  20. C (integration by parts)

Practice: Substitution (u-sub)

Evaluate each integral using substitution.
For each problem, a hint is provided — try first, then click to reveal if needed.

1. \[ \int 3x(2x^2+1)^5\,dx \]

Hint Let \(u=2x^2+1\). You’ll need to rewrite \(x\,dx\) in terms of \(du\).

2. \[ \int \frac{5x}{x^2+4}\,dx \]

Hint Let \(u=x^2+4\). Then \(du\) is proportional to \(x\,dx\).

3. \[ \int x^2 e^{x^3}\,dx \]

Hint Let \(u=x^3\). Then \(du\) contains \(x^2\,dx\).

4. \[ \int \frac{2x}{\sqrt{x^2+9}}\,dx \]

Hint Let \(u=x^2+9\). Then the integrand becomes a power of \(u\).

5. \[ \int \cos(4x)\,dx \]

Hint Let \(u=4x\). Don’t forget the constant when converting \(dx\).

6. \[ \int (x+1)^7\,dx \]

Hint Let \(u=x+1\). Then apply the power rule in \(u\).

7. \[ \int \frac{6x+9}{x^2+3x+1}\,dx \]

Hint Let \(u=x^2+3x+1\). Notice that \(6x+9 = 3(2x+3)\).

8. \[ \int x\sin(x^2)\,dx \]

Hint Let \(u=x^2\). Then \(du\) is proportional to \(x\,dx\).

9. \[ \int \frac{1}{(3x-2)^4}\,dx \]

Hint Rewrite as \((3x-2)^{-4}\). Let \(u=3x-2\).

10. \[ \int \sqrt{5x+1}\,dx \]

Hint Let \(u=5x+1\). Then rewrite the root as a power of \(u\).
Click to reveal answer key

Answer Key — Substitution

1. Let \(u=2x^2+1\), \(du=4x\,dx\), so \(x\,dx=\frac14 du\). \[ \int 3x(2x^2+1)^5dx=\frac34\int u^5\,du=\frac34\cdot\frac{u^6}{6}+C=\frac18u^6+C \] \[ \boxed{\frac{(2x^2+1)^6}{8}+C} \]

2. Let \(u=x^2+4\), \(du=2x\,dx\), so \(x\,dx=\frac12 du\). \[ \int \frac{5x}{x^2+4}dx=\frac52\int \frac{1}{u}\,du=\frac52\ln|u|+C \] \[ \boxed{\frac{5}{2}\ln(x^2+4)+C} \]

3. Let \(u=x^3\), \(du=3x^2\,dx\), so \(x^2\,dx=\frac13 du\). \[ \int x^2 e^{x^3}dx=\frac13\int e^u\,du=\frac13 e^u+C \] \[ \boxed{\frac13 e^{x^3}+C} \]

4. Let \(u=x^2+9\), \(du=2x\,dx\). \[ \int \frac{2x}{\sqrt{x^2+9}}dx=\int u^{-1/2}\,du=2u^{1/2}+C \] \[ \boxed{2\sqrt{x^2+9}+C} \]

5. Let \(u=4x\), \(du=4\,dx\), so \(dx=\frac14 du\). \[ \int \cos(4x)\,dx=\frac14\int \cos u\,du=\frac14\sin u+C \] \[ \boxed{\frac14\sin(4x)+C} \]

6. Let \(u=x+1\), \(du=dx\). \[ \int (x+1)^7dx=\int u^7\,du=\frac{u^8}{8}+C \] \[ \boxed{\frac{(x+1)^8}{8}+C} \]

7. Let \(u=x^2+3x+1\), \(du=(2x+3)\,dx\). Since \(6x+9=3(2x+3)\), \[ \int \frac{6x+9}{x^2+3x+1}dx = 3\int \frac{2x+3}{u}\,dx = 3\int \frac{1}{u}\,du = 3\ln|u|+C \] \[ \boxed{3\ln|x^2+3x+1|+C} \]

8. Let \(u=x^2\), \(du=2x\,dx\), so \(x\,dx=\frac12 du\). \[ \int x\sin(x^2)dx=\frac12\int \sin u\,du=-\frac12\cos u+C \] \[ \boxed{-\frac12\cos(x^2)+C} \]

9. Let \(u=3x-2\), \(du=3\,dx\), so \(dx=\frac13 du\). \[ \int (3x-2)^{-4}dx=\frac13\int u^{-4}\,du=\frac13\cdot\frac{u^{-3}}{-3}+C \] \[ \boxed{-\frac{1}{9(3x-2)^3}+C} \]

10. Let \(u=5x+1\), \(du=5\,dx\), so \(dx=\frac15 du\). \[ \int \sqrt{5x+1}\,dx=\frac15\int u^{1/2}\,du=\frac15\cdot\frac{2}{3}u^{3/2}+C \] \[ \boxed{\frac{2}{15}(5x+1)^{3/2}+C} \]

Practice: Integration by Parts

Evaluate each integral using integration by parts.
For each problem, a hint is provided — try first, then click to reveal if needed.

1. \[ \int x e^{5x}\,dx \]

Hint Choose \(u=x\) since it simplifies when differentiated.

2. \[ \int x^2 e^{2x}\,dx \]

Hint Let \(u=x^2\). You will need integration by parts twice.

3. \[ \int \ln x \, dx \]

Hint Write this as \(\ln x \cdot 1\) and choose \(dv=dx\).

4. \[ \int x\ln x \, dx \]

Hint Let \(u=\ln x\) and \(dv=x\,dx\).

5. \[ \int x^2 \ln x \, dx \]

Hint Let \(u=\ln x\). The remaining integral becomes a polynomial.

6. \[ \int (\ln x)^2 \, dx \]

Hint Treat this as \((\ln x)^2\cdot 1\). Choose \(dv=dx\).

7. \[ \int x(\ln x)^2\,dx \]

Hint Let \(u=(\ln x)^2\). You will reduce the problem to \(\int x\ln x\,dx\).

8. \[ \int x\sin x\,dx \]

Hint Let \(u=x\), \(dv=\sin x\,dx\). Remember \(\int \sin x\,dx=-\cos x\).

9. \[ \int x\cos x\,dx \]

Hint Let \(u=x\), \(dv=\cos x\,dx\). Remember \(\int \cos x\,dx=\sin x\).

10. \[ \int x^2 e^{-x}\,dx \]

Hint Let \(u=x^2\). This problem requires two rounds of integration by parts.
Click to reveal answer key

Answer Key — Integration by Parts

1. \[ \boxed{\frac{e^{5x}}{25}(5x-1)+C} \]

2. \[ \boxed{e^{2x}\left(\frac{x^2}{2}-\frac{x}{2}+\frac{1}{4}\right)+C} \]

3. \[ \boxed{x\ln x-x+C} \]

4. \[ \boxed{\frac{x^2}{2}\ln x - \frac{x^2}{4}+C} \]

5. \[ \boxed{\frac{x^3}{3}\ln x - \frac{x^3}{9}+C} \]

6. \[ \boxed{x\big((\ln x)^2 - 2\ln x + 2\big)+C} \]

7. \[ \boxed{\frac{x^2}{2}(\ln x)^2-\frac{x^2}{2}\ln x+\frac{x^2}{4}+C} \]

8. \[ \boxed{-x\cos x+\sin x+C} \]

9. \[ \boxed{x\sin x+\cos x+C} \]

10. \[ \boxed{-e^{-x}(x^2+2x+2)+C} \]

Practice: Partial Fractions (Factorable Quadratics)

Evaluate each integral using partial fractions.
All denominators are quadratic and factorable (repeated factors may appear).
For each problem, a hint is provided — try first, then click to reveal if needed.

1. \[ \int \frac{6}{x^2-9}\,dx \]

Hint Factor \(x^2-9=(x-3)(x+3)\).

2. \[ \int \frac{5}{x^2+3x-10}\,dx \]

Hint Factor the denominator into \((x-2)(x+5)\), then use \(\frac{A}{x-2}+\frac{B}{x+5}\).

3. \[ \int \frac{2x+1}{x^2-x-6}\,dx \]

Hint Factor \(x^2-x-6\). Use \(\frac{A}{x-3}+\frac{B}{x+2}\) and match coefficients.

4. \[ \int \frac{3x-2}{x^2-4x+3}\,dx \]

Hint Factor \(x^2-4x+3=(x-1)(x-3)\), then solve for \(A,B\).

5. \[ \int \frac{4}{x(x-5)}\,dx \]

Hint Use \(\frac{A}{x}+\frac{B}{x-5}\).

6. \[ \int \frac{7x+1}{(x-2)(x+1)}\,dx \]

Hint Use \(\frac{A}{x-2}+\frac{B}{x+1}\) and match coefficients.

7. \[ \int \frac{3}{(x-1)^2}\,dx \]

Hint This is a repeated linear factor: rewrite as \(3(x-1)^{-2}\).

8. \[ \int \frac{x}{(x+2)^2}\,dx \]

Hint For a repeated factor, use
\(\displaystyle \frac{A}{x+2}+\frac{B}{(x+2)^2}\).

9. \[ \int \frac{2x+5}{(x-1)(x-4)}\,dx \]

Hint Use \(\frac{A}{x-1}+\frac{B}{x-4}\) and solve for \(A,B\).

10. \[ \int \frac{1}{x^2-x}\,dx \]

Hint Factor \(x^2-x=x(x-1)\), then decompose.
Click to reveal answer key

Answer Key — Partial Fractions (with brief notes)

1. \[ \int \frac{6}{x^2-9}\,dx \] Factor \(x^2-9=(x-3)(x+3)\) and decompose: \[ \frac{6}{(x-3)(x+3)}=\frac{1}{x-3}-\frac{1}{x+3}. \] \[ \boxed{\ln|x-3|-\ln|x+3|+C=\ln\left|\frac{x-3}{x+3}\right|+C} \]

2. \[ \int \frac{5}{x^2+3x-10}\,dx \] Factor \(x^2+3x-10=(x-2)(x+5)\) and decompose: \[ \frac{5}{(x-2)(x+5)}=\frac{5/7}{x-2}-\frac{5/7}{x+5}. \] \[ \boxed{\frac{5}{7}\ln|x-2|-\frac{5}{7}\ln|x+5|+C} \]

3. \[ \int \frac{2x+1}{x^2-x-6}\,dx \] Factor \(x^2-x-6=(x-3)(x+2)\) and decompose: \[ \frac{2x+1}{(x-3)(x+2)}=\frac{7/5}{x-3}+\frac{3/5}{x+2}. \] \[ \boxed{\frac{7}{5}\ln|x-3|+\frac{3}{5}\ln|x+2|+C} \]

4. \[ \int \frac{3x-2}{x^2-4x+3}\,dx \] Factor \(x^2-4x+3=(x-1)(x-3)\) and decompose: \[ \frac{3x-2}{(x-1)(x-3)}=\frac{-1/2}{x-1}+\frac{7/2}{x-3}. \] \[ \boxed{-\frac12\ln|x-1|+\frac{7}{2}\ln|x-3|+C} \]

5. \[ \int \frac{4}{x(x-5)}\,dx \] Decompose: \[ \frac{4}{x(x-5)}=\frac{-4/5}{x}+\frac{4/5}{x-5}. \] \[ \boxed{-\frac45\ln|x|+\frac45\ln|x-5|+C} \]

6. \[ \int \frac{7x+1}{(x-2)(x+1)}\,dx \] Decompose: \[ \frac{7x+1}{(x-2)(x+1)}=\frac{5}{x-2}+\frac{2}{x+1}. \] \[ \boxed{5\ln|x-2|+2\ln|x+1|+C} \]

7. \[ \int \frac{3}{(x-1)^2}\,dx \] Rewrite as \(3(x-1)^{-2}\): \[ \boxed{-\frac{3}{x-1}+C} \]

8. First decompose: \[ \frac{x}{(x+2)^2} = \frac{1}{x+2} - \frac{2}{(x+2)^2} \]

Then integrate: \[ \boxed{\ln|x+2|+\frac{2}{x+2}+C} \]

9. \[ \int \frac{2x+5}{(x-1)(x-4)}\,dx \] Decompose: \[ \frac{2x+5}{(x-1)(x-4)}=\frac{-7/3}{x-1}+\frac{13/3}{x-4}. \] \[ \boxed{-\frac{7}{3}\ln|x-1|+\frac{13}{3}\ln|x-4|+C} \]

10. \[ \int \frac{1}{x^2-x}\,dx \] Factor \(x^2-x=x(x-1)\) and decompose: \[ \frac{1}{x(x-1)}=-\frac{1}{x}+\frac{1}{x-1}. \] \[ \boxed{-\ln|x|+\ln|x-1|+C=\ln\left|\frac{x-1}{x}\right|+C} \]