Question

Show that choosing a different anti-derivative $v+C$of $dv$will yield an equivalent formula for integration by parts,

as follows:

(a) Explain why what we need to show is that $uv-\int vdu=u\left(v+C\right)-\int \left(v+C\right)du$.

(b) Rewrite the equation from part (a), substituting$u=u\left(x\right)$, $v=v\left(x\right)$, and $du=u\text{'}\left(x\right)dx$.

(c) Prove the equality you wrote down in part (b).

Short answer

Part (a) We have added the constant term, which was eliminated during the derivative.

Part (b) $u\left(x\right)v\left(x\right)-\int v\left(x\right)u\text{'}\left(x\right)dx=u\left(x\right)\left(v\left(x\right)+C\right)-\int \left(v\left(x\right)+C\right)u\text{'}\left(x\right)dx$

Part (c) Proved in the step.

Step-by-step solution

Part (a) Step 1. Explain the result.

Consider the formula of by part,

$\int udv=uv-\int vdu$

Substitute $v=v+C$in the above integral.

$uv-\int vdu=u\left(v+C\right)-\int \left(v+C\right)du$

In the above by part method, by use v+C in place of v because we are doing integration and take the first function in terms of v. As the constant term C is added because the derivative of constant is zero.

Part 2. Step 1. Explanation.

Substitute the given values into the given equation.

$u\left(x\right)v\left(x\right)-\int v\left(x\right)u\text{'}\left(x\right)dx=u\left(x\right)\left(v\left(x\right)+C\right)-\int \left(v\left(x\right)+C\right)u\text{'}\left(x\right)dx$

Part (c) Step 1. Explanation.

Prove the equality you wrote down in part (b).

let y be a function,

$y=uv$

Do the derivative of the functions.

$\begin{array}{l}\frac{dy}{dx}=\frac{d\left(uv\right)}{dx}\\ =u\frac{dv}{dx}+v\frac{du}{dx}\end{array}$

Rearrange the rule to solve it.

$u\frac{dv}{dx}=\frac{d\left(uv\right)}{dx}-v\frac{du}{dx}$

Integrate both sides.

$\begin{array}{l}\int u\frac{dv}{dx}dx=\int \frac{d\left(uv\right)}{dx}dx-\int v\frac{du}{dx}dx\\ \int udv=uv-v\int du\end{array}$