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Romashka-Z-Leto [24]
3 years ago
14

Use integration by parts to find the integrals in Exercise. x^3 ln x dx.

Mathematics
1 answer:
34kurt3 years ago
5 0

Answer:

\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C.

Step-by-step explanation:

We have been given an indefinite integral \int \:x^3\:ln\:x\:dx. We are asked to find the value of the integral using integration by parts.

\int\: u\text{dv}=uv-\int\: v\text{du}

Let u=\text{ln}(x), v'=x^3.

Now, we will find du and v as shown below:

\frac{du}{dx}=\frac{d}{dx}(\text{ln}(x))

\frac{du}{dx}=\frac{1}{x}

du=\frac{1}{x}dx

v=\frac{x^{3+1}}{3+1}=\frac{x^{4}}{4}

Upon substituting our values in integration by parts formula, we will get:

\int \:x^3\:\text{ln}\:(x)\:dx=\text{ln}(x)*\frac{x^4}{4}-\int\: \frac{x^4}{4}*\frac{1}{x}dx

\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\int\: \frac{x^3}{4}dx

\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}\int\: x^3dx

\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}*\frac{x^{3+1}}{3+1}+C

\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}*\frac{x^4}{4}+C

\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C

Therefore, our required integral would be \frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C.

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