# Integration by Parts

# Integration by Parts

**Integration by Parts**

We have already seen the **reverse chain rule**. **Integration by parts** is the** reverse product rule**.

**Integration by parts** has many uses, most notably** integrating** things of the form x^{n}f(x).

For some questions, you need to **integrate by parts** **more than once** to get a result.

**Integration by Parts**

**Recall:** The **product rule**.

\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}

**Integration by parts** is the** reverse** of this.

{\LARGE \int}u\dfrac{dv}{dx}=uv-{\LARGE \int}v\dfrac{du}{dx}

You should choose u to be **something that differentiates nicely**, so that you can evaluate the **integral** on the** right-hand side**.

**Example: **Find \int x\sin(x)dx

u=x\;\;\;\dfrac{dv}{dx}=\sin(x)

\dfrac{du}{dx}=1\;\;\;v=-\cos(x)

\begin{aligned}\int x\sin(x)dx&=-x\cos(x)-\int-\cos(x)dx\\[1.2em]&=-x\cos(x)+\int \cos(x)dx\\[1.2em]&=-x\cos(x)+\sin(x)+c\end{aligned}

A Level**Integral of the Natural Logarithm**

One of the most important uses of **integration by parts** is that it gives us a way to find \int \ln(x)dx, by writng \ln(x) as

1\times \ln(x).

u=\ln(x)\;\;\;\dfrac{dv}{dx}=1

\dfrac{du}{dx}=\dfrac{1}{x}\;\;\;v=x

\begin{aligned}\int \ln(x)dx&=x\ln(x)-\int x\dfrac{1}{x}dx\\[1.2em]&=x\ln(x)-\int1dx\\[1.2em]&=x\ln(x)-x+c\\[1.2em]&=x(\ln(x)-1)+c\end{aligned}

A Level**Integrating by Parts Multiple Times**

Some **integrals** require you to **integrate by parts** **multiple times** to get a solution to.

**Example: **Find \int x^{2}e^{x}dx

u=x^{2}\;\;\;\dfrac{dv}{dx}=e^{x}

\dfrac{du}{dx}=2x\;\;\;v=e^{x}

\begin{aligned}\int x^{2}e^{x}dx=x^{2}e^{x}-\int2xe^{x}dx\end{aligned}

We must **integrate by parts** again the second **integral**.

u=2x\;\;\;\dfrac{dv}{dx}=e^{x}

\dfrac{du}{dx}=2\;\;\;v=e^{x}

\begin{aligned}\int x^{2}e^{x}dx&=x^{2}e^{x}-2xe^{x}+\int2e^{x}dx\\[1.2em]&=x^{2}e^{x}-2xe^{x}+2e^{x}+c\end{aligned}

A Level## Example Questions

**Question 1: **What is \int 2xe^{3x}dx?

**[2 marks]**

\dfrac{du}{dx}=2\;\;\;v=\dfrac{1}{3}e^{3x}

\begin{aligned}\int 2xe^{3x}dx&=\dfrac{2}{3}xe^{3x}-\int\dfrac{2}{3}e^{3x}dx\\[1.2em]&=\dfrac{2}{3}xe^{3x}-\dfrac{2}{9}e^{3x}+c\end{aligned}

**Question 2: **Find \int 4\ln(5x)dx

**[2 marks]**

Have that \int \ln(x)dx=x(\ln(x)-1)+c

The 4 will remain at the front, and the 5 inside the function becomes a \dfrac{1}{5} at the front when integrating.

\begin{aligned}\int 4\ln(5x)dx&=4\times\dfrac{1}{5}\times 5x(\ln(5x)-1)+c\\[1.2em]&=4x(\ln(5x)-1)+c\end{aligned}

**Question 3: **Find \int x^{3}\cos(2x)dx

**[4 marks]**

\dfrac{du}{dx}=3x^{2}\;\;\;v=\dfrac{1}{2}\sin(2x)

\begin{aligned}\int x^{3}\cos(2x)dx=\dfrac{1}{2}x^{3}\sin(2x)-\int\dfrac{3}{2}x^{2}\sin(2x)dx\end{aligned}

Integrate the integral on the right-hand side by parts.

u=\dfrac{3}{2}x^{2}\;\;\;\dfrac{dv}{dx}=\sin(2x)

\dfrac{du}{dx}=3x\;\;\;v=-\dfrac{1}{2}\cos(2x)

\begin{aligned}\int x^{3}\cos(2x)dx&=\dfrac{1}{2}x^{3}\sin(2x)+\dfrac{3}{4}x^{2}\cos(2x)-\int\dfrac{3}{2}x\cos(2x)dx\end{aligned}

Use integration by parts again.

u=\dfrac{3}{2}x\;\;\;\dfrac{dv}{dx}=\cos(2x)

\dfrac{du}{dx}=\dfrac{3}{2}\;\;\;v=\dfrac{1}{2}\sin(2x)

\begin{aligned}\int x^{3}\cos(2x)dx&=\dfrac{1}{2}x^{3}\sin(2x)+\dfrac{3}{4}x^{2}\cos(2x)-\dfrac{3}{4}x\sin(2x)+\int\dfrac{3}{4}\sin(2x)dx\\[1.2em]&=\dfrac{1}{2}x^{3}\sin(2x)+\dfrac{3}{4}x^{2}\cos(2x)-\dfrac{3}{4}x\sin(2x)-\dfrac{3}{8}\cos(2x)+c\end{aligned}

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