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3 years ago

Need help fast! need the value of x and please show the work!

Mathematics

1 answer:

lions [1.4K]3 years ago

6 0

By the external angle of a triangle theorem

m < ABD = m < BDC + m < BCD

That is 5x + 2 = x^2 - 1 + 4x - 3

x^2 + 4x - 5x - 1 - 3 - 2 = 0

x^2 - x - 6 = 0

(x - 3)(x + 2) = 0

x = 3, -2

x cannot be -2 because that would make the external angle, 5x-2 negative.

Answer x = 3

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2 years ago

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3 years ago

Integrate 1 - x / x(x2 + 1) d x by partial fractions.

solniwko [45]

Answer:

$log x-\frac{log(x^{2}+1) }{2}-tan^{-1} x$

Step-by-step explanation:

step 1:- by using partial fractions

$[tex]\frac{1-x}{x(x^{2}+1) } =\frac{A(x^{2}+1)+(Bx+C)(x }{x(x^{2}+1) }$ ......(1)

step 2:-

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substitute x =0 value in equation (2)

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comparing x co-efficient on both sides (in equation 2)

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step 3:-

substitute A,B,C values in equation (1)

now

$\\\int\limits^ {} \, \frac{1-x}{x(x^{2}+1) } d x =\int\limits^ {} \frac{1}{x} d x +\int\limits^ {} \frac{-x}{x^{2}+1 } d x -\int\limits \frac{1}{x^{2}+1 } d x$

by using integration formulas

i) by using $\int\limits \frac{1}{x} d x =log x+c........(a)\\\int\limits \frac{f^{1}(x) }{f(x)} d x= log(f(x)+c\\$ .....(b)

$\int\limits tan^{-1}x dx =\frac{1}{1+x^{2} } +C$ .....(c)

step 4:-

by using above integration formulas (a,b,and c)

we get answer is

$log x-\frac{log(x^{2}+1) }{2}-tan^{-1} x$

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