Free Printable WorksheetMathCalculus IIMath CurriculumAP Calculus BC Unit 6; FL MA.912.C.4.*; CCSS.MP1, MP6, MP7, MP8; Integrated Calculus II Standards CAL2-INT; TEKS 19 TAC section 111.55

CAL2.23 Single Step Integration by Parts Worksheet

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Worksheet Overview & Skills Covered

Core academic competencies and curriculum alignment

This Calculus II Math worksheet provides structured practice on CAL2.23 Single Step Integration by Parts. Students are guided through targeted exercises that build foundational understanding, computational accuracy, and critical thinking skills.

CAL2.23 Single-Step Integration by Parts - Select the correct antiderivative after one integration-by-parts step. Standards tags: AP Calculus BC Unit 6; FL MA.912.C.4.*; CCSS.MP1, MP6, MP7, MP8; Integrated Calculus II Standards CAL2-INT; TEKS 19 TAC section 111.55.

Whether you are an educator preparing classroom handouts, a tutor reinforcing weekly topics, or a parent supporting homeschool learning, this printable resource gives students the deliberate practice required to achieve subject fluency and master state standards.

Grade

Calculus II

Subject

Math

Domain

Math Curriculum

Formats

PDF & DOCX

Sample Practice Problems

Preview the types of problems and questions covered in this worksheet

Below are sample questions representative of the exercises included on this worksheet and related CAL2.23 Single Step Integration by Parts skill assessments:

Problem 1Difficulty: Mastery

Before evaluating any remaining integral, which expression is obtained from one integration-by-parts step on \(\int x² e^{3 x} \cos{(2 x )}\,dx\) using \(u=x² \cos{(2 x )}\) and \(dv=e^{3 x}\,dx\)?

A\(x² e^{3 x} \cos{(2 x )} - \int 2 x (- x \sin{(2 x )} + \cos{(2 x )}) e^{3 x}\,dx + C\)

B\(\frac{x² e^{3 x} \cos{(2 x )}}{3} - \int \frac{2 x (- x \sin{(2 x )} + \cos{(2 x )}) e^{3 x}}{3}\,dx + C\)

C\(\frac{x² e^{3 x} \cos{(2 x )}}{3} - \int \frac{x² e^{3 x} \cos{(2 x )}}{3}\,dx + C\)

D\(\frac{x² e^{3 x} \cos{(2 x )}}{3} + \int \frac{2 x (- x \sin{(2 x )} + \cos{(2 x )}) e^{3 x}}{3}\,dx + C\)

Solution & Explanation:Since \(v=\frac{e^{3 x}}{3}\) and \(du=2 x (- x \sin{(2 x )} + \cos{(2 x )})\,dx\), substitution into \(\int u\,dv=uv-\int v\,du\) gives \(\frac{x² e^{3 x} \cos{(2 x )}}{3} - \int \frac{2 x (- x \sin{(2 x )} + \cos{(2 x )}) e^{3 x}}{3}\,dx + C\). The most common errors are changing the minus sign before the remaining integral or using dv in place of its antiderivative v.

Problem 2Difficulty: Mastery

A student chooses \(u=x² \sin{(x )}\) and \(dv=e^-x\,dx\) for \(\int x² e^-x \sin{(x )}\,dx\). Select the expression that follows from \(uv-\int v\,du\) after exactly one step.

A\(- x² e^-x \sin{(x )} + \int x (x \cos{(x )} + 2 \sin{(x )}) e^-x\,dx + C\)

B\(x² e^-x \sin{(x )} - \int x (x \cos{(x )} + 2 \sin{(x )}) e^-x\,dx + C\)

C\(- x² e^-x \sin{(x )} - \int x (x \cos{(x )} + 2 \sin{(x )}) e^-x\,dx + C\)

D\(- x² e^-x \sin{(x )} + \int x² e^-x \sin{(x )}\,dx + C\)

Solution & Explanation:Since \(v=- e^-x\) and \(du=x (x \cos{(x )} + 2 \sin{(x )})\,dx\), substitution into \(\int u\,dv=uv-\int v\,du\) gives \(- x² e^-x \sin{(x )} + \int x (x \cos{(x )} + 2 \sin{(x )}) e^-x\,dx + C\). The most common errors are changing the minus sign before the remaining integral or using dv in place of its antiderivative v.

Problem 3Difficulty: Mastery

For \(\int x e^{2 x} \arctan{(x )}\,dx\), take \(u=x \arctan{(x )}\) and \(dv=e^{2 x}\,dx\). Which one-step integration-by-parts result is correct? Assume the interval stays in the domain of the functions shown.

A\(\frac{x e^{2 x} \arctan{(x )}}{2} + \int \frac{(x + (x² + 1) \arctan{(x )}) e^{2 x}}{2 (x² + 1)}\,dx + C\)

B\(\frac{x e^{2 x} \arctan{(x )}}{2} - \int \frac{(x + (x² + 1) \arctan{(x )}) e^{2 x}}{2 (x² + 1)}\,dx + C\)

C\(x e^{2 x} \arctan{(x )} - \int \frac{(x + (x² + 1) \arctan{(x )}) e^{2 x}}{x² + 1}\,dx + C\)

D\(\frac{x e^{2 x} \arctan{(x )}}{2} - \int \frac{x e^{2 x} \arctan{(x )}}{2}\,dx + C\)

Solution & Explanation:Since \(v=\frac{e^{2 x}}{2}\) and \(du=\frac{x + (x² + 1) \arctan{(x )}}{x² + 1}\,dx\), substitution into \(\int u\,dv=uv-\int v\,du\) gives \(\frac{x e^{2 x} \arctan{(x )}}{2} - \int \frac{(x + (x² + 1) \arctan{(x )}) e^{2 x}}{2 (x² + 1)}\,dx + C\). The most common errors are changing the minus sign before the remaining integral or using dv in place of its antiderivative v.

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QWhat grade level and standards does this worksheet align with?

This worksheet is designed specifically for Calculus II Math students. It aligns with educational standard AP Calculus BC Unit 6; FL MA.912.C.4.*; CCSS.MP1, MP6, MP7, MP8; Integrated Calculus II Standards CAL2-INT; TEKS 19 TAC section 111.55 (Select the correct antiderivative after one integration-by-parts step. Standards tags: AP Calculus BC Unit 6; FL MA.912.C.4.*; CCSS.MP1, MP6, MP7, MP8; Integrated Calculus II Standards CAL2-INT; TEKS 19 TAC section 111.55.).