Question

. Evaluate
2x+1
dx
Vx2-3x+1​

Answer 1

Answer:

Step-by-step explanation:

math]3 x^3 - x^2 + 2 x - 4 = 3 x^3 - 3 x^2 + 2 x^2 - 2 x + 4 x - 4[/math]

[math]=3 x^2 (x - 1) + 2 x (x - 1) + 4 (x - 1) = (x - 1) (3 x^2 + 2 x + 4)[/math]

[math]x^2 - 3 x + 2 = x^2 - 2 x - x + 2 = x (x - 2) - 1 (x - 2) = (x - 1) (x - 2)[/math]

[math]\displaystyle\int_{0}^{1} \dfrac{(3 x^3 - x^2 + 2 x - 4)}{(x^2 - 3 x + 2)}\,dx[/math]

[math]=\displaystyle\int_{0}^{1} \dfrac{(x - 1) (3 x^2 + 2 x + 4)}{(x - 1) (x - 2)}\,dx[/math]

[math]=\displaystyle\int_{0}^{1} \dfrac{(3 x^2 + 2 x + 4)}{(x - 2)}\,dx[/math]

[math]=\displaystyle\int_{0}^{1} \dfrac{(3 x^2 - 6 x + 8 x + 4)}{(x - 2)}\,dx[/math]

[math]=\displaystyle\int_{0}^{1} \dfrac{(3 x^2 - 6 x)}{(x - 2)}\,dx + \displaystyle\int_{0}^{1} \dfrac{(8 x + 4)}{(x - 2)}\,dx[/math]

[math]=\displaystyle\int_{0}^{1} \dfrac{3 x(x - 2)}{(x - 2)}\,dx + 8\displaystyle\int_{0}^{1} \dfrac{(x + 0.5)}{(x - 2)}\,dx[/math]

[math]=\displaystyle\int_{0}^{1} 3 x\,dx + 8\displaystyle\int_{0}^{1} \dfrac{(x - 2 + 2.5)}{(x - 2)}\,dx[/math]

[math]=\left.\dfrac{3 x^2}{2}\right|_{0}^{1} + 8\displaystyle\int_{0}^{1} \dfrac{(x - 2)}{(x - 2)}\,dx + 8\displaystyle\int_{0}^{1} \dfrac{(2.5)}{(x - 2)}\,dx[/math]

[math]=\dfrac{3}{2} + 8\displaystyle\int_{0}^{1}\,dx + 20\displaystyle\int_{0}^{1} \dfrac{1}{(x - 2)}\,dx[/math]

[math]=\dfrac{3}{2} + \left. 8 x\right|_{0}^{1} + \left. 20 \ln |x - 2|\right|_{0}^{1}[/math]

[math]=\dfrac{3}{2} + 8 - 20\ln (2)[/math]

[math]= \dfrac{19}{2} - 20\ln (2)[/math]

[math]=-4.3629[/math]

Answer 2

Answer:

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Step-by-step explanation:

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