Original formula =-∫ TDT/√ (t 4+ 1)

=- 1/2*∫d(t^2)/√[(t^2)^2+ 1]

=- 1/2*ln|t^2+√(t^4+ 1)|+c

=- 1/2*ln| 1/x^2+√( 1/x^4+ 1)|+c

2. let x = sindx = costdt.

Original formula =∫costdt/(sint+cost)

Let a = ∫ costdt/(Sint+cost) b = ∫ Sintdt/(Sint+cost).

a+B =∫(Sint+cost)dt/(Sint+cost)= t+c 1

A-B=∫ (cost -Sint)dt/(Sint+ cost) =∫d(Sint+ cost) /(Sint+ cost) = ln | Sint+ cost |+C2.

A=(t+ln|sint+cost|)/2+C

So the original formula = (arcsinx+ln | x+√ (1-x 2) |)/2+C.

3. The original formula = ∫ [1+√ x-√ (1+x)] dx/(1+2 √ x+x-1-x)

= 1/2*∫x^(- 1/2)+ 1-√( 1+x)/√x dx

=√x+x/2- 1/2 *∫√( 1+x)dx/√x

Let t = √ x t2 = x1+x = t2+1dx = 2tdt.

Original formula = √ x+x/2-∫√ (t 2+ 1) dt.

=√x+x/2-t/2*√(t^2+ 1)- 1/2*ln|t+√(t^2+ 1)|+c

=√x+x/2- 1/2*√(x^2+x)- 1/2*ln|√x+√(x+ 1)|+c

4. Let t = lnxx = e t dx = e tdt.

Original formula = ∫√ (1+t) dt/t.

Let u = √ (1+t) t = u 2-1dt = 2udu.

Original formula = 2 * ∫ u 2du/(u 2- 1)

) du

= 2 *∫du+∫du/(u- 1)-∫du/(u+ 1)

= 2u+ln | u- 1 |-ln | u+ 1 |+C

= 2 √( 1+t)+ln | √( 1+t)- 1 |-ln | √( 1+t)+ 1 |+C

= 2 √( 1+lnx)+ln | √( 1+lnx)- 1 |-ln | √( 1+lnx)+ 1 |+C