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N76 [4]
2 years ago
8

The gradient of the line y = x+3y=9​

Mathematics
1 answer:
Karolina [17]2 years ago
4 0

Answer:

change the expression to y

Step-by-step explanation:

x + 3y = 9

3y =  - x + 9

y =  \frac{ - 1}{3} x +  \frac{9}{3}

the gradient is the coefficient of x,

-  \frac{1}{3}

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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)

<u>step 2:-</u>

solving on both sides

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substitute x =0 value in equation (2)

1=A(1)+0

<u>A=1</u>

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0 = A+B

0 = 1+B

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

<u>-</u>1  =  C

<u>step 3:-</u>

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)

<u>step 4:-</u>

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

we get answer is

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

6 0
3 years ago
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