Question

∫ 5tanx / (tanx-2) dx

Answer 1

let u = tanx - 2 
du/dx = sec²x 
dx = du/sec²x 

u+2 = tanx 
(u+2)² = tan²x 
(u+2)² + 1 = tan²x + 1 
(u+2)² + 1 = sec²x 
u² + 4u + 5 = sec²x 

∫ 5tanx / (tanx-2) dx 
= ∫ 5(u+2)/u ∙ du/sec²x 
= ∫ (5u+10)/u ∙ du/(u² + 4u + 5) 
= ∫ (5u+10) / [u(u² + 4u + 5)] du 

Partial fraction decomposition: 
(5u+10) / [u(u² + 4u + 5)] = A/u + (Bu+C) / (u² + 4u + 5) 
Multiply by u(u² + 4u + 5): 
5u+10 = A(u² + 4u + 5) + u(Bu+C) 
5u+10 = Au² + Bu² + 4Au + Cu + 5A 

Matching up corresponding terms, 
Au²+Bu² = 0u² since no quadratic term on the L.H.S. 

4Au + Cu = 5u 
4A + C = 5 

10 = 5A 

A = 2 
B = -2 
C = -3 

∴ ∫ (5u+10) / [u(u² + 4u + 5)] du 
= ∫2/u du - ∫(2u+3) / (u² + 4u + 5) du 
= 2ln|u| - ∫(2u+4) / (u² + 4u + 5) du + ∫ 1 du / (u² + 4u + 5) 
= 2ln|u| - ln|u² + 4u + 5| + ∫ 1 du / ((u + 2)² + 1) 
= 2ln|u| - ln|u² + 4u + 5| + arctan(u + 2) 
= 2ln|tanx-2| - ln(sec²x) + arctan(tanx-2+2) 
= 2ln|tanx-2| - 2ln(secx) + arctan(tanx) 
= 2 ln|tanx-2)/secx)| + x 
= 2ln|sinx - 2cosx| + x + C

Answer 2

Step-by-step explanation:

The solution of this integral is given above .