Question

don't post irrelevant answers

Answer 1

∫ [ 1 / 3x² +13x - 10 ] dx

= ∫ [ 1 / 3x² + 15x - 2x - 10 ] dx

= ∫ [ 1 / 3x( x + 5) - 2(x + 5) ] dx

= ∫ [ 1 / ( x+ 5) ( 3x - 2) ] dx

= ∫ [ 1 / ( x+ 5) ( 3x - 2) ] dx

= ∫ [ (1/ 17 ) × {3 ( x+ 5) - ( 3x - 2) } / ( x+ 5) ( 3x - 2) ] dx

= (1/17) ∫ [ 3 / ( 3x - 2) ] dx - (1/17) ∫ [ 1 / ( x + 5 ) ] dx

Let

y = 3x - 2 => dy = 3dx

z = x+5 => dz = dx

So

Given Expression

= (1/17) ∫ ( 1 / y) dy - (1/17) ∫ [ 1 / z ) ] dz

= (1/17) log |y| - (1/17) log |z| + c

= (1/17) log | 3x - 2 | - (1/17) log | x+ 5 | + c

Answer 2

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