Katie's copy of Integration by parts

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Question 1

Find the following indefinite integral.

$I=\displaystyle \int \simplify[std]{x*({b}x+{c})^{m}} dx$
You are given that $I=\simplify[std]{({b}x+{c})^{m+1}/{b^2*(m+1)*(m+2)}*g(x)+C}$
For a polynomial $g(x)$ . You have to find $g(x)$ .

$g(x)=\;$ interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{15 x + 7}$

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The formula for integrating by parts is
$\int u\frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx.$
Also you need to know that for $n \neq -1$ :
$\int (ax+b)^n dx = \frac{1}{a(n+1)}(ax+b)^{n+1}+C$

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Advice

The formula for integrating by parts is

$\int u\frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx.$

We choose $u = x$ and $\displaystyle \frac{dv}{dx} = \simplify[std]{({b}*x+{c})^{m}}$ .

So $\displaystyle \frac{du}{dx}$ = $1$ and $\displaystyle v = \simplify[std]{(1/{(m+1)*b})*({b}*x+{c})^{m+1}}$ .

Hence,
$\begin{eqnarray*} \displaystyle \int \simplify[std]{x*({b}x+{c})^{m}} dx &=& uv - \int v \frac{du}{dx} dx \\ &=& \simplify[std]{(x/{(m+1)*b})*({b}*x+{c})^{m+1} - (1/{(m+1)*b})*Int (({b}*x+{c})^{m+1}, x)} \\ &=& \simplify[std]{(x/{(m+1)*b})*({b}*x+{c})^{m+1} - (1/{(m+1)*(m+2)*b^2})*({b}*x+{c})^{m+2}+C} \\ &=&\simplify[std]{({b}*x+{c})^{m+1}/{(m+1)*(m+2)*b^2}*({b*(m+2)}x - ({b}x+{c}))+C}\\ &=&\simplify[std]{({b}*x+{c})^{m+1}/{(m+1)*(m+2)*b^2}*({b*(m+1)}x - {c})+C} \end{eqnarray*}$
The solution is: $\simplify[std]{g(x)={b*(m+1)}*x-{c}}$ .

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Katie's copy of Integration by parts

Advice