MHB What are the steps to simplify this integral?

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To simplify the integral $$\int \frac{x^2}{x^2 + 9} \,dx$$, the expression can be rewritten as $$\frac{x^2 + 9 - 9}{x^2 + 9}$$, which simplifies to $$1 - \frac{9}{x^2 + 9}$$. This allows the integral to be separated into two parts: $$\int dx - 9\int \frac{1}{x^2 + 9} \,dx$$. The first integral is straightforward, while the second requires knowledge of the arctangent function. Following these steps leads to a clearer approach for solving the integral.
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I have this integral:

$$\int_{}^{} \frac{x^2}{x^2 + 9} \,dx$$

And I'm trying to simplify it to:

$$\int_{}^{}\,dx - 9\int_{}^{} \frac{1}{x^2 + 9}\,dx$$

But I'm not sure of the steps necessary to do this.
 
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You could look at it this way:

$$\frac{x^2}{x^2+9}=\frac{x^2+9-9}{x^2+9}=1-\frac{9}{x^2+9}$$
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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