Question

Find the indefinite integral. (Note: Solve by the simplest method—not all require integration by parts. Use C for the constant of integration.)
\int 9arctan x dx

\int 9 arctan x dx

Answer 1

Answer:

$\int{9 \arctan{\left(x \right)} d x} = 9 x \arctan{\left(x \right)} - \frac{9}{2} \ln{\left(\left|{x^{2} + 1}\right| \right)}+C$

Step-by-step explanation:

To find $\int \:9\arctan \left(x\right)dx$ you must:

Step 1: Apply the constant multiple rule $\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$ with c = 9 and $f{\left(x \right)} = \arctan {\left(x \right)}$

$\int{9 \arctan }{\left(x \right)} d x}} =9 \int{\arctan {\left(x \right)} d x}$

Step 2: For the integral $\int{\arctan}{\left(x \right)} d x}$, use integration by parts $\int {u} {dv} ={u}{v} - \int {v}{du}$

Let ${u}={\arctan}{\left(x \right)}$ and $dv=dx$.

Then

${du}=\left({\arctan}{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x^{2} + 1}$ and ${v}=\int{1 d x}=x$

The integral can be rewritten as

$9 {\int{\arctan}{\left(x \right)} d x}}=9 {\left(\arctan{\left(x \right)} \cdot x-\int{x \cdot \frac{1}{x^{2} + 1} d x}\right)}=9{\left(x\arctan{\left(x \right)} - \int{\frac{x}{x^{2} + 1} d x}\right)}$

Let $u=x^{2} + 1$

Then $du=\left(x^{2} + 1\right)^{\prime }dx = 2 x dx$ and we have that $x dx = \frac{du}{2}$.

Therefore,

$9 x \arctan{\left(x \right)} - 9 {\int{\frac{x}{x^{2} + 1} d x}} = 9 x \arctan{\left(x \right)} - 9 {\int{\frac{1}{2 u} d u}}$

$9 x \arctan{\left(x \right)} - 9 {\int{\frac{1}{2 u} d u}} = 9 x \arctan{\left(x \right)} - 9 {\left(\frac{1}{2} \int{\frac{1}{u} d u}\right)}$

Step 3: The integral of $\frac{1}{u}$ is $\int{\frac{1}{u} d u} = \ln{\left(u \right)}$

$x \arctan{\left(x \right)} - \frac{9}{2} {\int{\frac{1}{u} d u}} = 9 x \arctan{\left(x \right)} - \frac{9}{2} {\ln{\left(u \right)}}$

Step 4: Recall that $u=x^{2} + 1$

$9 x \arctan{\left(x \right)} - \frac{9}{2} \ln{\left({u} \right)} = 9 x \arctan{\left(x \right)} - \frac{9}{2} \ln{\left({\left(x^{2} + 1\right)} \right)}$

Therefore,

$\int{9 \arctan{\left(x \right)} d x} = 9 x \arctan{\left(x \right)} - \frac{9}{2} \ln{\left(\left|{x^{2} + 1}\right| \right)}$

Step 5: Add the constant of integration

$\int{9 \arctan{\left(x \right)} d x} = 9 x \arctan{\left(x \right)} - \frac{9}{2} \ln{\left(\left|{x^{2} + 1}\right| \right)}+C$

Answer 2

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)