Question

Integrate 1/x from - 1 to - 4 w.r.t.x ​

Answer 1

Answer:

your solution is in above pic

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Answer 2

Concept:-

Inorder to solve this problem, one must be aware about a basic formula of integration. We will use the below formula to solve the given question.

$\boxed{\sf\int \dfrac{1}{x} dx = \log(|x|) + C}$

Don't forget to add the modulus for value of x :)

Note that,

$\boxed{ \begin{array}{l} \rm \dfrac{d}{dx} \log (- x) = \dfrac{1}{ - x} \cdot - 1 = \bf \blue{ \dfrac{1}{x}} \\ \\ \rm \dfrac d{dx} \log(x) = \bf \blue{\dfrac{1}{x}} \end{array}}$

Solution:-

Given definite integral,

$\displaystyle \implies \int _{ - 4}^{- 1} \frac{1}{x} dx$

Applying the formula, we get:

$\displaystyle \implies \log( |x| ) \bigg |^{ - 1} _{ - 4}$

$\displaystyle \implies \log( | - 1| ) - \log( | - 4|)$

$\displaystyle \implies \log(1) - \log(4)$

$\displaystyle \implies 0- \log(4)$

$\displaystyle \implies - \log(4)$

$\displaystyle \implies \approx - 1.3863$

Hence the required answer is:

$\underline {\boxed{\int _{ - 4}^{- 1} \frac{1}{x} dx = - \log(4)}}$

Additional Information:-

$\boxed{\begin{array}{l} \textbf{ \textsf{Basic integration formulas: }}\\ \\ \circ\sf\:\int{1\:dx}=x+c\\\\\circ\sf\:\int{a\:dx}=ax+c\\\\\circ\sf\:\int{x^n\:dx}=\dfrac{x^{n+1}}{n+1}+c\\\\\circ\sf\:\int{sin\:x\:dx}=-cos\:x+c\\\\\circ\sf\:\int{cos\:x\:dx}=sin\:x+c\\\\\circ\sf\:\int{sec^2x\:dx}=tan\:x+c\\\\\circ\sf\:\int{e^x\:dx}=e^x+c\end{array}}$