Evaluate : (int_{1}^{3} (frac{x – 1}{(x + 1)^{2}}) mathrm {d} x).
Explanation
(int_{1}^{3} (frac{x – 1}{(x + 1)^{2}}) mathrm {d} x)
Let (x + 1 = u, x – 1 = u – 2)
When x = 3, u = 3 + 1 = 4
x = 1, u = 1 + 1 = 2
(therefore int_{1}^{3} (frac{x – 1}{(x + 1)^{2}}) mathrm {d} x equiv int_{2}^{4} (frac{u – 2}{u^{2}}) mathrm {d} u)
= (int_{2}^{4} (frac{u}{u^{2}} – frac{2}{u^{2}}) mathrm {d} u)
= (int_{2}^{4} (frac{1}{u} – 2u^{-2}) mathrm {d} u)
= ([ln u – frac{2u^{-1}}{-1}]|_{2}^{4})
= ([ln u + 2u^{-1}]|_{2}^{4})
= ([ln u + frac{2}{u}]|_{2}^{4})
= ((ln 4 + frac{2}{4}) – (ln 2 + frac{2}{2}))
= (ln 4 – ln 2 – frac{1}{2})
= (ln (frac{4}{2}) – frac{1}{2})
= (ln 2 – frac{1}{2}).