Question

How to solve this?
\int \frac { 4 - 3 x ^ { 2 } } { ( 3 x ^ { 2 } + 4 ) ^ { 2 } } d x​

Answer 1

`\Large \mathbb{SOLUTION:}`

`\begin{array}{l} \displaystyle \int (4 - 3x^2)/((3x^2 + 4)^2) dx \\ \\ = \displaystyle \int (4 - 3x^2)/(x^2\left(3x + (4)/(x)\right)^2) dx \\ \\ = \displaystyle \int ((4)/(x^2) - 3)/(\left(3x + (4)/(x)\right)^2) dx \\ \\ \text{Let }u = 3x + (4)/(x) \implies du = \left(3 - (4)/(x^2)\right)\ dx \\ \\ \text{So the integral becomes} \\ \\ = \displaystyle -\int (du)/(u^2) \\ \\ = -(u^(-2 + 1))/(-2 + 1) + C \\ \\ = (1)/(u) + C \\ \\ = (1)/(3x + (4)/(x)) + C \\ \\ = \boxed{(x)/(3x^2 + 4) + C}\end{array}`