Question
Evaluate the following integrals:
$\int_{0}^\limits{1}\frac{24\text{x}^3}{(1+\text{x}^2)^4}\text{ dx}$
$\int_{0}^\limits{1}\frac{24\text{x}^3}{(1+\text{x}^2)^4}\text{ dx}$
✓
Answer
Let $1+\text{x}^2=\text{t}$
Differentiating w.r.t. x, we get
$2\text{xdx}=\text{dt}$
Now, $\text{x}=0\Rightarrow\text{t}=1$
$\text{x}=1\Rightarrow\text{t}=2$
$\int_{0}^\limits{1}\frac{24\text{x}^3}{(1+\text{x}^2)^4}\text{ dx}=\int_{1}^\limits{2}\frac{12(\text{t}-1)}{\text{t}^4}\text{ dt}$
$=12\int_{1}^\limits{2}\Big(\frac{1}{\text{t}^3}-\frac{1}{\text{t}^4}\Big)\text{dt}$
$=12\Big[-\frac{1}{2\text{t}^2}-\frac{1}{3\text{t}^3}\Big]^2_1$
$=12\Big[-\frac{1}{8}+\frac{1}{24}+\frac{1}{2}-\frac{1}{3}\Big]$
$=12\Big[\frac{-3+1+12-8}{24}\Big]$
$=\frac{12\times2}{24}=1$
Differentiating w.r.t. x, we get
$2\text{xdx}=\text{dt}$
Now, $\text{x}=0\Rightarrow\text{t}=1$
$\text{x}=1\Rightarrow\text{t}=2$
$\int_{0}^\limits{1}\frac{24\text{x}^3}{(1+\text{x}^2)^4}\text{ dx}=\int_{1}^\limits{2}\frac{12(\text{t}-1)}{\text{t}^4}\text{ dt}$
$=12\int_{1}^\limits{2}\Big(\frac{1}{\text{t}^3}-\frac{1}{\text{t}^4}\Big)\text{dt}$
$=12\Big[-\frac{1}{2\text{t}^2}-\frac{1}{3\text{t}^3}\Big]^2_1$
$=12\Big[-\frac{1}{8}+\frac{1}{24}+\frac{1}{2}-\frac{1}{3}\Big]$
$=12\Big[\frac{-3+1+12-8}{24}\Big]$
$=\frac{12\times2}{24}=1$
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