Q. 0 - Calculus

Question

Read the section and make your own summary of the material.

Step-by-Step Solution

Answer

The formula for integration by parts: If $u=u\left(x\right)$ and $v=v\left(x\right)$ are differentiable functions, then $\int u dv = uv - \int v du$

Step 1. Given information

Section, 5.2 Integration by parts of the chapter, 5. Techniques of integration.

Step 2. Summary of the section

• Reversing the Product Rule: If $u$ and $v$ are functions such that $u\text{'}\left(x\right)v\left(x\right)+u\left(x\right)v\text{'}\left(x\right)$ is integrable, then $\int \left(u\text{'}\left(x\right)v\left(x\right)+u\left(x\right)v\text{'}\left(x\right)\right) dx = u\left(x\right)v\left(x\right) + C$
• The formula for integration by parts: If $u=u\left(x\right)$ and $v=v\left(x\right)$ are differentiable functions, then $\int u dv = uv - \int v du$
• It is best to try algebraic simplification and u-substitution before attempting integration by parts.
• Choose $u$ and $dv$ so that the associated $du$ and $v$ are simpler than what we started with.
• Integral of the Natural Logarithm Function: $\int \ln x dx = x \ln x - x + C$
• Integrals of Inverse Sine and Inverse Tangent: $$\begin{gathered} \int {\sin }^{-1}x dx = x {\sin }^{-1}x + \sqrt{1-x^2} + C \\ \int {\tan }^{-1}x dx = x {\tan }^{-1}x - \frac{\ln \left(x^2+1\right)}{2} + C \end{gathered}$$
• Integration by Parts for Definite Integrals: If $u=u\left(x\right)$ and $v=v\left(x\right)$ are differentiable functions on $\left[a, b\right]$, then ${\int }_a^{b}u dv = {\left[\int u dv\right]}_a^b={\left[uv\right]}_a^b-{\int }_a^{b}v du$