1 (4 points): 4 3 2 1 0 Evaluate the following integral, without using indefinite integrals. Find and simplify an exact answer

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Mathematics Linh Đan 6 months 2021-07-23T14:46:50+00:00 2021-07-23T14:46:50+00:00 1 Answers 7 views 0

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Captcha* Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )

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hongcuc2 · Answer 1 · 2021-07-23T14:48:17+00:00

hongcuc2

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2021-07-23T14:48:17+00:00 July 23, 2021 at 2:48 pm

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Answer:

$\mathbf{ 12 In (4)- \dfrac{27}{4} }$

Step-by-step explanation:

Suppose the given integral is:

$\mathtt{I = \int ^5_2 x In(x-1)dx}$

By utilizing integration by parts method:

$\mathbf{I = \Big[In (x-1)\int xdx\Big ]^5_2-\int^5_2\Big(\dfrac{d}{dx}(In(x-1) \int x dx \Big)dx}$

$\mathbf{\implies \Big[ \dfrac{x^2}{2}In (x-1)\Big ]^5_2-\int^5_2\Big(\dfrac{1}{x-1}\Big)\dfrac{x^2}{2} \Big)dx}$

$= \mathbf{ \Big[ \dfrac{5^2}{2}In (5-1)- \dfrac{2^2}{2}In (2-1)\Big ]-\dfrac{1}{2} \int ^5_2 \dfrac{x^2-1+1}{x-1} dx}$

$= \mathbf{ \dfrac{25}{2}In (4)- \dfrac{2^2}{2}In (2-1)\Big ]-\dfrac{1}{2} \int ^5_2 \dfrac{x^2-1+1}{x-1} dx}$

$= \mathbf{ \dfrac{25}{2}In (4)- \dfrac{1}{2}\Big \ \int ^5_2\Big( \dfrac{x^2-1}{x-1}+\dfrac{1}{x-1} \Big) dx}$

$= \mathbf{ \dfrac{25}{2}In (4)- \dfrac{1}{2}\Big \ \int ^5_2\Big( {x+1}+\dfrac{1}{x-1} \Big) dx}$

$= \mathbf{ \dfrac{25}{2}In (4)- \dfrac{1}{2} \Big [\dfrac{x^2}{2}+x+In(x-1) \Big] }^5_2$

$= \mathbf{ \dfrac{25}{2}In (4)- \dfrac{1}{2} \Big [\dfrac{5^2}{2}+5+In(5-1)-\dfrac{2^2}{2}-2-In(2-1) \Big] }$

$= \mathbf{ \dfrac{25}{2}In (4)- \dfrac{1}{2} \Big [\dfrac{25}{2}+5 -4+In(4)\Big] }$

$= \mathbf{ \dfrac{25}{2}In (4)- \dfrac{1}{2} \Big [\dfrac{25}{2}+1+In(4)\Big] }$

$= \mathbf{ 12 In (4)- \dfrac{27}{4} }$