1C ∫ √tan(𝑥) 𝑑𝑥 1 − 𝑠𝑒𝑛2 (𝑥) U=tan(x) du= 𝑠𝑒𝑐 2 (𝑥)𝑑𝑥 𝑑𝑢 = 𝑑𝑥 𝑠𝑒𝑐 2 (𝑥) √U 1 √U 𝑐𝑜𝑠2 (𝑥) =∫ 2 ∗ 2( ) 𝑑𝑢 𝑐𝑜𝑠 (𝑥) 𝑠𝑒𝑐 𝑥 =∫ 2 ∗ 𝑐𝑜𝑠 (𝑥) =∫ 1 𝑑𝑢 √U 𝑑𝑢 1 3 2𝑢2 = 3 +𝑐 2√𝑡𝑎𝑛3 (𝑥) = 3 +𝑐 2C ∫(𝑥 2 − 2𝑥 + 5)𝑒 −𝑥 𝑑𝑥 u =𝑥 2 − 2𝑥 + 5 dv=𝑒 −𝑥 du=(2x-2)dx v=−𝑒 −𝑥 = −(𝑥 2 − 2𝑥 + 5)𝑒 −𝑥 + ∫(2𝑥 − 2)𝑒 −𝑥 𝑑𝑥 u=2x-2 dv=𝑒 −𝑥 du=2dx v=−𝑒 −𝑥 = −(𝑥 2 − 2𝑥 + 5)𝑒 −𝑥 − (2𝑥 − 2)𝑒 −𝑥 + ∫ 2𝑒 −𝑥 𝑑𝑥 = −𝑥 2 𝑒 −𝑥 + 2𝑥𝑒 −𝑥 − 5𝑒 −𝑥 − 2𝑥𝑒 −𝑥 + 2𝑒 −𝑥 + ∫ 2𝑒 −𝑥 𝑑𝑥 = −𝑥 2 𝑒 −𝑥 − 3𝑒 −𝑥 + 2 ∫ 𝑒 −𝑥 𝑑𝑥 = −𝑥 2 𝑒 −𝑥 − 3𝑒 −𝑥 − 2𝑒 −𝑥 + 𝑐 = 𝑥 2 𝑒 −𝑥 − 5𝑒 −𝑥 + 𝑐 3c ∫ 𝑥2 √𝑥 2 − 4 𝑑𝑥 Solución ∫ 𝑥2 √𝑥 2 − 4 𝑑𝑥 X=2sec(a) dx= 2sec(a)tan(a)da =∫ (2sec(a))2 √(2sec(a))2 − 4 =∫ =∫ ∗ 2sec(a)tan(a)da 8𝑠𝑒𝑐 3 (𝑎) tan(𝑎) √4(sec 2 (𝑎) − 1) 8𝑠𝑒𝑐 3 (𝑎) tan(𝑎) √4(tan2 (𝑎) 𝑑𝑎 𝑑𝑎 8𝑠𝑒𝑐 3 (𝑎) tan(𝑎) =∫ 𝑑𝑎 2 tan(𝑎) = 4 ∫ 𝑠𝑒𝑐 3 (𝑎) 𝑑𝑎 = 4 ∫ sec(𝑎) (𝑠𝑒𝑐 2 (𝑎))𝑑𝑎 = 4 ∫ sec(𝑎) (𝑡𝑎𝑛2 (𝑎) + 1)𝑑𝑎 = 4 ∫ sec(𝑎) (𝑡𝑎𝑛2 (𝑎))𝑑𝑎 + 4 ∫ sec(𝑎) 𝑑𝑎 Entonces 4 ∫ sec(𝑎) (𝑡𝑎𝑛2 (𝑎))𝑑𝑎 1 𝑠𝑒𝑛(𝑎) 2 = 4∫ ∗ ( ) 𝑑𝑎 cos(𝑎) cos(𝑎) = 4 ∫ 𝑠𝑒𝑛(𝑎) ∗ 𝑠𝑒𝑛(𝑎) 𝑑𝑎 𝑐𝑜𝑠 3 (𝑎) U=sen(a) 𝑑𝑣 = 𝑠𝑒𝑛(𝑎) 𝑐𝑜𝑠 3 (𝑎) du= cos(a)da 𝑣 = = 1 2𝑐𝑜𝑠 2 (𝑎) 2𝑠𝑒𝑛(𝑎) 1 − 2 ∫ ∗ cos(𝑎) 𝑑𝑎 𝑐𝑜𝑠 2 (𝑎) 𝑐𝑜𝑠 2 (𝑎) = = 𝑑𝑎 2𝑠𝑒𝑛(𝑎) − 2 ∫ sec(𝑎) 𝑑𝑎 𝑐𝑜𝑠 2 (𝑎) 2𝑠𝑒𝑛(𝑎) − 2 ln(sec(𝑎) + tan(𝑎)) + 𝑐 𝑐𝑜𝑠 2 (𝑎) Entonces 4 ∫ sec(𝑎) 𝑑𝑎 = 4 ln(sec(𝑎) + tan(𝑎)) + 𝑐 Por lo tanto ∫ 𝑥2 √𝑥 2 − 4 𝑑𝑥 = 2𝑠𝑒𝑛(𝑎) − 2 ln(sec(𝑎) + tan(𝑎)) + 4 ln(sec(𝑎) + tan(𝑎)) + 𝑐 𝑐𝑜𝑠 2 (𝑎) = 2𝑠𝑒𝑛(𝑎) + 2 ln(sec(𝑎) + tan(𝑎)) + 𝑐 𝑐𝑜𝑠 2 (𝑎) Reemplazando a 𝑥√𝑥 2 − 4 𝑥 √𝑥 2 − 4 = + 2 ln ( + )+𝑐 2 2 2 4c∞ 0 4 𝑑𝑥 2 −∞ 4 + 𝑥 ∫ 0 4 𝑑𝑥 2 4 + 𝑥 −𝑛 = lim ∫ 𝑛→∞ = lim ∫ 𝑛→∞ 4 𝑥 2 4 (( ) + 1) 2 = lim ∫ 𝑛→∞ 1 𝑥 2 ((2) + 1) 𝑥 𝑢= 2 𝑑𝑥 𝑑𝑥 2𝑑𝑢 = 𝑑𝑥 = lim (2 ∫ 𝑛→∞ 1 𝑑𝑢) (𝑢2 + 1) = lim (2 arctan(𝑢)) 𝑛→∞ 𝑥 = lim (2 arctan ( ) 𝑛0𝑙 ) 𝑛→∞ 2 0 𝑛 = lim (2 arctan ( ) − 2arctan( ) 𝑛→∞ 2 2 𝑛 = lim − 2arctan( ) 𝑛→∞ 2 𝑛 = −2 lim arctan ( ) 𝑛→∞ 2 𝑛 = −2𝑎𝑟𝑐𝑡𝑎𝑛 ( lim ) 𝑛→∞ 2 𝜋 = −2 arctan(+∞) = −2 = −𝜋 2