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# 1.1: Integration by parts

[ "stage:draft", "article:topic", "Integration by Parts", "authorname:thangarajahp" ]

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Integration by parts (IBP)

$\int\,u \,dv= u\,v-\int\, v\, du.$

Note

When deciding which function should be u and dv, it is best to choose u as the function that will simplify easily.

The priority for choosing u is:

1. $$u = ln x$$

2. $$u = x^n$$,  where n= integer

3. $$u = e^{nx}$$, where n= integer

Example $$\PageIndex{1}$$:

Find $$\int x^2 \, e^{-x}\, dx$$.

Answer:

Using the priority listed above, let $$u=x^2$$ and $$dv= e^{-x} \,dx$$.

Since $$u=x^2$$  ,              $$du=2x$$ .

Since   $$dv= e^{-x}$$,        $$v= \int e^{-x} \, dx). \(= -e^{-x}$$

Using the IBP,

$$\int\,u \,dv= u\,v-\int\, v\, du$$

$$= x^2\, e^{-x}\ -\int\, e^{-x}\, 2x\, dx$$

$$= ln x\, \dfrac{x^2}{2}\ - \dfrac{1}{2}\int\ x\, dx$$

$$= ln x\, \dfrac{x^2}{2}\ - \dfrac{1}{2}\, \dfrac{x^2}{2}$$

$$= ln x\, \dfrac{x^2}{2}\ - \dfrac{x^2}{4}$$

Simplifying the answer,

$$\int x \,lnx \, dx\, = \dfrac{x^2}{2}\, (ln x\, -\, \dfrac{1}{2}\, ) +\, C$$

Reminder: Make sure to write + C at the end of the answer as C is a constant.

Example $$\PageIndex{2}$$:

Find $$\int x \,lnx \, dx$$.

Answer:

Using the priority listed above, let $$u=ln x$$ and $$dv=x\,dx$$.

Since $$u=ln x$$  ,      $$du=\dfrac{1}{x}$$ .

Since   $$dv=x$$,        $$v= \dfrac{x^2}{2}$$.

Using the IBP,

$$\int\,u \,dv= u\,v-\int\, v\, du$$

$$= ln x\, \dfrac{x^2}{2}\ -\int\, \dfrac{x^2}{2}\, x\, dx$$

$$= ln x\, \dfrac{x^2}{2}\ - \dfrac{1}{2}\int\ x\, dx$$

$$= ln x\, \dfrac{x^2}{2}\ - \dfrac{1}{2}\, \dfrac{x^2}{2}$$

$$= ln x\, \dfrac{x^2}{2}\ - \dfrac{x^2}{4}$$

Simplifying the answer,

$$\int x \,lnx \, dx\, = \dfrac{x^2}{2}\, (ln x\, -\, \dfrac{1}{2}\, ) +\, C$$

Reminder: Make sure to write + C at the end of the answer as C is a constant.

Example $$\PageIndex{3}$$:

Find $$\int \, lnx \, dx$$.

Answer:

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Exercise $$\PageIndex{1}$$

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