IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

Greatdreams

Let (1 + x) = z ⁶

So dx = 6 z ⁵

So it becomes,

6 z ⁵/ z³ - z²

= 6 z³ / z - 1

= 6 [(z³ - 1) + 1]/ z- 1

= 6 (z³ - 1)/(z-1) + 1/z-1

= 6 (z-1)(z² + z + 1) / (z-1) + 1/z-1

= 6 (z² + z + 1) + 1/(z-1)

So

6 ( z ² + z + 1) dz +

dz/(z-1)

= 6 z³/3 + 6 z²/2 + 6 z + ln ( z - 1)

= 2 z ³ + 3 z² + 6 z + ln (z - 1)

= 2 (1 + x)^3/6 + 3 ( 1 + x)^2/6 + 6 ( 1 + x)^1/6 + ln [ (1 + x) ^1/6 - 1] + c

= 2

( 1 + x) + 3 ( 1 + x) ^1/3 + 6(1 + x)^1/6+ ln [ (1 + x) ^1/6 - 1 ] + c

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