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Question: How do you integrate \(\int{{{x}^{2}}\ln x dx}\) by integration by parts method?...

How do you integrate ∫x2ln⁡xdx\int{{{x}^{2}}\ln x dx} by integration by parts method?

Explanation

Solution

We first discuss the integration by parts method. Integration by parts method is usually used for the multiplication of the functions and their integration. We take two arbitrary functions to express the theorem. We take the u=ln⁡x,v=x2u=\ln x,v={{x}^{2}} for our integration ∫x2ln⁡xdx\int{{{x}^{2}}\ln x dx}. We use the formulas ∫xndx=xn+1n+1+c\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c, ddx(ln⁡x)=1x\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}.

Complete step-by-step solution:
We need to find the integration of ∫x2ln⁡xdx\int{{{x}^{2}}\ln x dx} using integration by parts method.
Integration by parts method is usually used for the multiplication of the functions and their integration.
Let’s assume f(x)=g(x)h(x)f\left( x \right)=g\left( x \right)h\left( x \right). We need to find the integration of ∫f(x)dx=∫g(x)h(x)dx\int{f\left( x \right)dx}=\int{g\left( x \right)h\left( x \right)dx}.
We take u=g(x),v=h(x)u=g\left( x \right),v=h\left( x \right). This gives ∫f(x)dx=∫uvdx\int{f\left( x \right)dx}=\int{uvdx}.
The theorem of integration by parts gives ∫uvdx=u∫vdx−∫(dudx∫vdx)dx\int{uvdx}=u\int{vdx}-\int{\left( \dfrac{du}{dx}\int{vdx} \right)dx}.
For our integration ∫x2ln⁡xdx\int{{{x}^{2}}\ln x dx}, we take u=ln⁡x,v=x2u=\ln x,v={{x}^{2}}. This is done as per the ILATE rule which helps us decide which function to choose as u and v. So, the order for choosing u is inverse, logarithmic, algebra, trigonometric, exponential.
Now we complete the integration ∫x2ln⁡xdx=ln⁡x∫x2dx−∫(d(ln⁡x)dx∫x2dx)dx\int{{{x}^{2}}\ln x dx}=\ln x\int{{{x}^{2}}dx}-\int{\left( \dfrac{d\left( \ln x \right)}{dx}\int{{{x}^{2}}dx} \right)dx}.
We have the differentiation formula for u=ln⁡xu=\ln x where dudx=ddx(ln⁡x)=1x\dfrac{du}{dx}=\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}.
The integration formula for ∫xndx=xn+1n+1+c\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c.
We apply these formulas to complete the integration and get
∫x2ln⁡xdx=ln⁡x(x33)−∫(1x×x33)dx=ln⁡x(x33)−13∫x2dx\int{{{x}^{2}}\ln x dx}=\ln x\left( \dfrac{{{x}^{3}}}{3} \right)-\int{\left( \dfrac{1}{x}\times \dfrac{{{x}^{3}}}{3} \right)dx}=\ln x\left( \dfrac{{{x}^{3}}}{3} \right)-\dfrac{1}{3}\int{{{x}^{2}}dx}.
We have one more integration part remaining. To complete that we use ∫xndx=xn+1n+1+c\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c
So, ∫x2ln⁡xdx=ln⁡x(x33)−13∫x2dx=ln⁡x(x33)−x39+c\int{{{x}^{2}}\ln x dx}=\ln x\left( \dfrac{{{x}^{3}}}{3} \right)-\dfrac{1}{3}\int{{{x}^{2}}dx}=\ln x\left( \dfrac{{{x}^{3}}}{3} \right)-\dfrac{{{x}^{3}}}{9}+c. Here cc is the integral constant.
We simplify the form to get

& \int{{{x}^{2}}\ln x dx} \\\ & =\dfrac{{{x}^{3}}}{9}\left( 3\ln x-1 \right)+c \\\ \end{aligned}$$ Therefore, the integration by parts of $$\int{{{x}^{2}}\ln x dx}$$ gives $$\dfrac{{{x}^{3}}}{9}\left( 3\ln x-1 \right)+c$$. **Note:** In case one of two functions is missing and we need to form the by parts method, we will take the multiplying constant 1 as the second function. For example: if we need to find $$\int{\ln x dx}$$, we have only one function. So, we take constant 1 as the second function where $u=\ln x,v=1$. But we need to remember that we won’t perform by parts by taking $u=1$.