Answer:
∫▒〖arctan(x).1 dx=arctan(x).x〗-1/2 ln(1+x^2 )+C
Step-by-step explanation:
∫▒〖1st .2nd dx=1st∫▒〖2nd dx〗-∫▒〖(derivative of 1st) dx∫▒〖2nd dx〗〗〗
Let 1st=arctan(x)
And 2nd=1
∫▒〖arctan(x).1 dx=arctan(x) ∫▒〖1 dx〗-∫▒〖(derivative of arctan(x))dx∫▒〖1 dx〗〗〗
As we know that
derivative of arctan(x)=1/(1+x^2 )
∫▒〖1 dx〗=x
So
∫▒〖arctan(x).1 dx=arctan(x).x〗-∫▒〖(1/(1+x^2 ))dx.x〗…………Eq1
Let’s solve ∫▒(1/(1+x^2 ))dx by substitution now
Let 1+x^2=u
du=2xdx
Multiply and divide ∫▒〖(1/(1+x^2 ))dx.x〗 by 2 we get
1/2 ∫▒〖(2/(1+x^2 ))dx.x〗=1/2 ∫▒(2xdx/u)
1/2 ∫▒(2xdx/u) =1/2 ∫▒(du/u)
1/2 ∫▒(2xdx/u) =1/2 ln(u)+C
1/2 ∫▒(2xdx/u) =1/2 ln(1+x^2 )+C
Putting values in Eq1 we get
∫▒〖arctan(x).1 dx=arctan(x).x〗-1/2 ln(1+x^2 )+C (required soultion)
Hey there!!
In order to solve this equation, we will have to use the distributive property.
What is distributive property?
We will need to distribute one term to the other terms preset.
Example : 4 ( 3x - 1 )
sing distributive property, we will have to distribute the term outside the parenthesis to the terms inside. In this case, we will have to distribute 4 to 3x and -1
( 4 × 3x ) + ( 4 × -1 )
12x + ( -4 )
= 12x - 4
This is after implementing the distributive property.
Now moving back onto the question
4 ( x + 5 ) = 3 ( x - 2 ) - 2 ( x + 2 )
Let us first solve 4 ( x + 5 )
Distribute 4 to x and 5
( 4 × x ) + ( 4 × 5 )
4x + 20
Now let's solve 3 ( x - 2 )
Distribute 3 to x and -2
( 3 × x ) + ( 3 × - 2 )
3x - 6
Solve for -2 ( x + 2 )
Distribute -2 to x and 2
( -2 × x ) + ( -2 × 2 )
-2x - 4
Now, let's get everything back together
4x + 20 = 3x - 6 - 2x - 4
Combine all the like terms
4x + 20 = x - 10
Adding 10 on both sides
4x + 20 + 10 = x - 10 + 10
4x + 30 = x
Subtracting 4x on both sides
4x - 4x + 30 = x - 4x
30 = -3x
Dividing by -3 on both sides
30 / -3 = -3x / -3
- 10 = x
<h2>x = - 10 </h2><h3>Hope my answer helps!</h3>
Answer:
There could be many but one of them is
(m+5)(m-9)=0....
Answer:
(sss) condition
Step-by-step explanation:
hope the ans helped you
Answer
Find out the conversion factor for seconds to minutes and convert 135 seconds to minutes.
To prove
1 minute = 60 second
for seconds to minutes.

Therefore the conversion factor for seconds to minutes be

Now convert 135 seconds to minutes.

= 2.25 minutes
Hence proved