Question

the value of ( x^3 long ×dx is?​

Answer 1

EXPLANATION.

$\sf \implies \displaystyle \int x^{3} log(x) dx$

As we know that,

In this type of question we can apply Integration by parts, we get.

If u and v are two functions of x then,

$\sf \implies \displaystyle \int(u . v)dx = u \bigg(\int v dx \bigg) \ - \int \bigg(\dfrac{du}{dx} . \bigg( \int v dx \bigg) \bigg) dx.$

From the first letter of the word,

I = inverse trigonometric functions.

L = Logarithmic functions.

A = Algebraic Functions.

T = Trigonometric functions.

E = Exponential functions.

We get a word = ILATE.

First arrange the functions in the order according to letters of this word and then integrate by parts, we get.

x³ = Algebraic functions. = 2nd functions.

㏒(x) = Logarithmic functions = 1st functions.

$\sf \implies \displaystyle \bigg(log(x)\bigg) \int x^{3}dx \ - \int \bigg[ \bigg(\dfrac{d(logx)}{dx} \int x^{3}dx\bigg) \bigg]dx$

$\sf \implies \displaystyle \bigg(log(x) \bigg) \ . \dfrac{x^{4} }{4} \ - \int \bigg[ \dfrac{1}{x} \times \dfrac{x^{4} }{4} \bigg]dx.$

$\sf \implies \displaystyle \dfrac{x^{4}(log x) }{4} \ - \dfrac{1}{4} \int x^{3} dx.$

$\sf \implies \displaystyle \dfrac{x^{4} (logx)}{4} \ - \dfrac{1}{4} \bigg(\dfrac{x^{4} }{4} \bigg) + C.$

$\sf \implies \displaystyle \dfrac{x^{4} (logx)}{4} \ - \dfrac{x^{4} }{16} + C.$

$\sf \implies \displaystyle \int x^{3} log(x)dx = \displaystyle \dfrac{x^{4} (logx)}{4} \ - \dfrac{x^{4} }{16} + C.$