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Question
Solution
\[Let\ I = \int_0^1 \frac{24 x^3}{\left( 1 + x^2 \right)^4} d\ x . Then, \]
\[Let\ x^2 = t . Then, 2x\ dx = dt\]
\[When\ x = , t = 0\ and\ x = 1, t = 1\]
\[ \therefore I = \int_0^1 \frac{12t}{\left( 1 + t \right)^4} dt\]
\[\text{Integrating by parts}\]
\[I = 12 \left[ \frac{t}{- 3 \left( 1 + t \right)^3} \right]_0^1 + 12 \int_0^1 \frac{1}{3 \left( 1 + t \right)^3}dt\]
\[ \Rightarrow I = 12\left\{ \left[ \frac{t}{- 3 \left( 1 + t \right)^3} \right]_0^1 - \left[ \frac{1}{6 \left( 1 + t \right)^2} \right]_0^1 \right\}\]
\[ \Rightarrow I = 12\left\{ - \frac{1}{24} - 0 - \frac{1}{24} + \frac{1}{6} \right\}\]
\[ \Rightarrow I = 12 \times \frac{1}{12}\]
\[ \Rightarrow I = 1\]
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