Help meh plzzzzzzzzzzzz!!!!!!!!!

Please, I need help in this ??

nignag [31]

Answer:

$\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c$

Step-by-step explanation:

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

Adding and Subtracting 1 to the Numerator

$\int\frac{x^{4} - 1 + 1}{x^{4} -1}dx$

Dividing Numerator seperately by $x^{4} - 1$

$\int 1 + \frac{1}{x^{4}-1 }\, dx$

Here integral of 1 is x +c1 (where c1 is constant of integration

$x + c1 + \int\frac{1}{(x-1)(x+1)(x^{2}+1)}\, dx$ ----------------------------------(1)

We apply method of partial fractions to perform the integral

$\frac{1}{(x-1)(x+1)(x^{2}+1)}$ = $\frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{x^{2} + 1}$ ------------------------------------------(2)

$\frac{1}{(x-1)(x+1)(x^{2}+1)} = \frac{A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)}{(x-1)(x+1)(x^{2} +1)}$

1 = $A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)$ -------------------------(3)

Substitute x= 1 , -1 , i in equation (3)

1 = A(1+1)(1+1)

A = $\frac{1}{4}$

1 = B(-1-1)(1+1)

B = $-\frac{1}{4}$

1 = C(i-1)(i+1)

C = $-\frac{1}{2}$

Substituting A, B, C in equation (2)

$\int\frac{x^{4}}{x^{4} -1}dx$ = $\int\frac{1}{4(x-1)} - \frac{1}{4(x+1)} -\frac{1}{2(x^{2}+1) }$

On integration

Here $\int \frac{1}{x}dx = lnx and \int\frac{1}{x^{2}+1 } dx = arctanx$

$\int\frac{x^{4}}{x^{4} -1}dx$ = $\frac{1}{4} ln(x-1)$ - $\frac{1}{4} ln(x+1)$ - $\frac{1}{2} arctanx$ + c2---------------------------------------(4)

Substitute equation (4) back in equation (1) we get

$x + c1 + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1) - \frac{1}{2} arctanx + c2$

Here c1 + c2 can be added to another and written as c

Therefore,

$\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c$

4 0

4 years ago

Alliic

pochemuha

First, we need to solve the first equation in order to see what the options need to be equivalent to. our first step to solving this is to distribute -3 through the parenthesis
y + 5 = -3x + 3
next, move the constant to the right side and change its sign
y = -3x + 3 - 5
calculate the difference
y = -3x - 2
since we cant find x, your final solution for y is y = -3x - 2, x ∈ R
now that we know the solution to that equation, we can now solve the options listed to see which ones are equivalent.
first is y = -3x - 2, which is very simple to solve because y has already been found. since we know y and there is no solution for x, our answer to this equation is y = -3x -2, x ∈ R. since this is the exact same answer as our first equation, we know that the A is equivalent to y + 5 = -3(x - 1)
the next equation is also already solved for y and has no x, so the answer to this equation is y = -4x - 5, x ∈ R. since this is does not have the same answer that <span>y + 5 = -3(x - 1) had</span>, option B is not equivalent.
our next equation is 3x + y = -2 does not provide y for us, so we must solve it.
first, you need to move the variable to the right side and change its sign
3x = -2 - y
now divide both sides of the equation by 3
x = - 2/3 = y/3
write all the numerators above their common denominators
x = - 2 + y/3
since we dont know what y is, our final solution is
x = - $\frac{2+y}{3}$ , y ∈ R
since this answer does not match the answer we are looking for, option C is not equivalent.
finally, we must solve -4x + y = -5
the first step to solving this is to move the variable to the right side and change its sign
-4x = -5 - y
divide both sides of the equation by -4
x = 5/4 + y/4
now, write all the numerators above their common denominators
x = 5 + y/4
since we dont know what y is, our answer is x = $\frac{5+y}{4}$ , y ∈ R
since this is also not equivalent to <span>y + 5 = -3(x - 1), our only answer is option A.
let me know if you have any further questions
:)</span>

5 0

3 years ago

Find the coordinate of the other endpoint if the midpoint is (5,5) and the endpoint is (-4,1)

Dima020 [189]

Answer:

p(14,9)

Step-by-step explanation:

steps are in picture.

Note: if you have any question about it you can ask.