Question

Integration of logx/(1+logx)per power 2​

Answer 1

Answer:

x / ( 1 + logx ) +c

Step-by-step explanation:

To find --->

∫ logx / ( 1 + logx )² dx

Solution---> ∫ { logx / ( 1 + logx )² } dx

We add and subtract 1 in numerator

= ∫ { ( 1 + logx ) - 1 / ( 1 + logx )² } dx

Now seperate it in to two intregals

= ∫ {( 1 + logx ) / ( 1 + logx )² - 1 / ( 1 + logx )² } dx

We know that aᵐ ÷ aⁿ = aᵐ⁻ⁿ , applying it here we get,

= ∫ { ( 1 + logx )¹⁻² - ( 1 + logx )⁻² } dx

= ∫ ( 1 + logx)⁻¹ ( 1 ) dx - ∫ ( 1 + logx )⁻² dx

Now we apply intregation by parts in first intregal and keeping second integral unchanged , considering ( 1 + logx)⁻¹ as first function and 1 as second function.

We have some formulee

a) d/dx ( logx ) = 1 / x

b) d/dx ( xⁿ ) = nxⁿ⁻¹

c) ∫ 1 dx = x

Applying these formulee here,

= ( 1 + logx )⁻¹ ( x ) - ∫ (-1)(1 + logx)⁻¹⁻¹ ( 0 + 1/x) x dx

- ∫ ( 1 + logx )⁻² dx

= x/ ( 1 + logx ) + ∫ ( 1 + logx )⁻² dx - ∫ ( 1 + logx )⁻² dx

Both integral cancel out each other and we get answer

= x / ( 1 + logx ) + c

Answer 2

Answer:

click the above image for step by step solution.