Respuesta :
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)
Answer:
[tex]\int{9 \arctan{\left(x \right)} d x} = 9 x \arctan{\left(x \right)} - \frac{9}{2} \ln{\left(\left|{x^{2} + 1}\right| \right)}+C[/tex]
Step-by-step explanation:
To find [tex]\int \:9\arctan \left(x\right)dx[/tex] you must:
Step 1: Apply the constant multiple rule [tex]\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx[/tex] with c = 9 and [tex]f{\left(x \right)} = \arctan {\left(x \right)}[/tex]
[tex]\int{9 \arctan }{\left(x \right)} d x}} =9 \int{\arctan {\left(x \right)} d x}[/tex]
Step 2: For the integral [tex]\int{\arctan}{\left(x \right)} d x}[/tex], use integration by parts [tex]\int {u} {dv} ={u}{v} - \int {v}{du}[/tex]
Let [tex]{u}={\arctan}{\left(x \right)}[/tex] and [tex]dv=dx[/tex].
Then
[tex]{du}=\left({\arctan}{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x^{2} + 1}[/tex] and [tex]{v}=\int{1 d x}=x[/tex]
The integral can be rewritten as
[tex]9 {\int{\arctan}{\left(x \right)} d x}}=9 {\left(\arctan{\left(x \right)} \cdot x-\int{x \cdot \frac{1}{x^{2} + 1} d x}\right)}=9{\left(x\arctan{\left(x \right)} - \int{\frac{x}{x^{2} + 1} d x}\right)}[/tex]
Let [tex]u=x^{2} + 1[/tex]
Then [tex]du=\left(x^{2} + 1\right)^{\prime }dx = 2 x dx[/tex] and we have that [tex]x dx = \frac{du}{2}[/tex].
Therefore,
[tex]9 x \arctan{\left(x \right)} - 9 {\int{\frac{x}{x^{2} + 1} d x}} = 9 x \arctan{\left(x \right)} - 9 {\int{\frac{1}{2 u} d u}}[/tex]
[tex]9 x \arctan{\left(x \right)} - 9 {\int{\frac{1}{2 u} d u}} = 9 x \arctan{\left(x \right)} - 9 {\left(\frac{1}{2} \int{\frac{1}{u} d u}\right)}[/tex]
Step 3: The integral of [tex]\frac{1}{u}[/tex] is [tex]\int{\frac{1}{u} d u} = \ln{\left(u \right)}[/tex]
[tex]x \arctan{\left(x \right)} - \frac{9}{2} {\int{\frac{1}{u} d u}} = 9 x \arctan{\left(x \right)} - \frac{9}{2} {\ln{\left(u \right)}}[/tex]
Step 4: Recall that [tex]u=x^{2} + 1[/tex]
[tex]9 x \arctan{\left(x \right)} - \frac{9}{2} \ln{\left({u} \right)} = 9 x \arctan{\left(x \right)} - \frac{9}{2} \ln{\left({\left(x^{2} + 1\right)} \right)}[/tex]
Therefore,
[tex]\int{9 \arctan{\left(x \right)} d x} = 9 x \arctan{\left(x \right)} - \frac{9}{2} \ln{\left(\left|{x^{2} + 1}\right| \right)}[/tex]
Step 5: Add the constant of integration
[tex]\int{9 \arctan{\left(x \right)} d x} = 9 x \arctan{\left(x \right)} - \frac{9}{2} \ln{\left(\left|{x^{2} + 1}\right| \right)}+C[/tex]
