∫ 2 e x + 3 e - x 4 e x + 7 e - x d x = 1 14 u x + v log e 4 e x + 7 e - x + C , where C is a constant of integration, then u + v is equal to

Question

&#8747; 2 e x + 3 e - x 4 e x + 7 e - x d x = 1 14 u x + v log e 4 e x + 7 e - x + C , where C is a constant of integration, then u + v is equal to

Aicharya · Accepted Answer

I = &#8747; 2 e x + 3 e - x 4 e x + 7 e - x d x = &#8747; 2 e x + 3 4 e 2 x + 7 d x Here 2 e 2 x + 3 = A 8 e 2 x + B 4 e 2 x + 7 &#8658; 2 e 2 x + 3 = ( 8 A + 4 B ) e 2 x + 7 B &#8658; B = 3 7 &#160; &#38; &#160; A = 1 28 I = &#8747; 1 28 8 e 2 x + 3 7 4 e 2 x + 7 4 e 2 x + 7 d x I = 1 28 ln 4 e 2 x + 7 + 3 7 x + C I = 1 28 ln &#8739; e x 4 e x + 7 e - x &#8739; + 3 7 x + C &#8658; u = 13 2 &#38; v = 1 2 &#8658; u + v = 13 2 + 1 2 = 7

∫ 2 e x + 3 e - x 4 e x + 7 e - x d x = 1 14 u x + v log e 4 e x + 7 e - x + C , where C is a constant…

Solution

&#8747; 2 e x + 3 e - x 4 e x + 7 e - x d x = 1 14 u x + v log e 4 e x + 7 e - x + C , where C is a constant…

Solution

∫ 2 e x + 3 e - x 4 e x + 7 e - x d x = 1 14 u x + v log e 4 e x + 7 e - x + C , where C is a constant…