Hi I am trying to evaluate a integral and I am hoping for some assistance
$$\int \frac{9x^2}{(x^6+9)} dx $$
I have rewritten the problem as shown below
$$\int 9x^2(x^6+9)^{-1}dx$$
Therefore I attempted the question by using integration by parts
$$\int udv = uv- \int vdu$$
$$u = (x^6+9)^{-1}$$ $$du = -6x^5(x^6+9)^{-2} dx$$
$$dv = 9x^2dx$$ $$v = 3x^3$$
Therefore substituting into the integration by parts formula we get
$$\int 9x^2(x^6+9)^{-1} dx = 3x^3(x^6+1)^{-1} - \int -18x^8(x^6+9)^{-2} dx$$
$$\int 9x^2(x^6+9)^{-1} dx = 3x^3(x^6+1)^{-1} - \int 3x^3 \frac{-6x^5}{(x^6+9)^2}dx$$
Then i tried using integration by substitution here
$$let u = x^6+9$$
$$\frac{du}{dx} = 6x^5$$
$$\frac{du}{6x^5} = dx$$
$$\int 9x^2(x^6+1)^{-1} dx = 3x^3(x^6+1)^{-1} - \int \frac{-3x^3}{u^2}du$$
This is where i have reached in terms of evaluating the problem When making the substitution I have gotten a $3x^3$ within the integral I am hoping someone can help evaluate this integral.
Now i am working on Picards Method of successive approximations
$$y_0 = y_0 \int f(x,y_0) dx$$
the ODE given was $$y' = \frac{x^2}{y^2+1}$$
$$x_0 = 0, y_0 =0$$
Therefore first approximation given by the following
$$y_1 = y_0 + \int f(x,y_0)dx$$
therefore y1
$$y_1 = 0 + \int \frac{x^2}{0+1} dx$$
$$y_1 = \frac{x^3}{3}$$
Second approximation
$$y_2 = y_0 + \int f(x,y_1)dx$$
Therefore we get the following
$$y_2 = 0 + \int \frac{x^2}{\frac{x^6}{9}+1} dx$$
and this is why i am trying to evalaute the integral but the solution which i am seeing is in the form
$$y = \frac{x^3}{3}+\frac{-x^9}{81}$$
and i am trying to see how they arrvied at this answer
$$I=\int \frac{9x^2}{(x^6+9)} dx$$ $$I=\int \frac{3dx^3}{(x^3)^2+9} =\int \frac{dx^3/3}{(x^3/3)^2+1} $$ $$I=\int \frac{du}{u^2+1} $$ Where $u=\dfrac {x^3}3$. Then use the $\arctan $ function. $$I=\arctan u +C=\arctan \left(\dfrac {x^3}3 \right)+C $$