Math, asked by dhiraj48, 1 year ago

integrate x³-1/x³+x dx

Answers

Answered by amitnrw
3

Given  : (x³-1)/(x³+x) dx

To Find : Integrate

Solution:

(x³-1)/(x³+x)

Add and subtract x in numerator

= (x³ + x  - x - 1) /(x³ + x)

= (x³ + x)/(x³ + x) - (x + 1)/(x³ + x)

= 1 - (x + 1)/x(x²  + 1)

(x + 1)/x(x²  + 1) =  A/x  + (Bx +C)/(x²  + 1)

=> x + 1  = A(x²  + 1) + Bx² + Cx

x = 0 => 1 = A

x = - 1

=> 0 = 2(1) + B -  C => B - C = -2

x = 1

=> 2 = 2(1) + B + C  => B + C  = 0

=> B = - 1   , C = 1

(x + 1)/x(x²  + 1) =  1/x  + (-x +1)/(x²  + 1)

(x³-1)/(x³+x)  = 1 - ( 1/x  + (-x +1)/(x²  + 1) )

= 1  - 1/x  + x/(x²+ 1)  - 1/(x² + 1)

∫(x³-1)/(x³+x) dx

= ∫dx - ∫dx/x  + ∫xdx/(x² + 1)  - ∫dx/(x² + 1)

= ∫dx - ∫dx/x  + (1/2)∫2xdx/(x² + 1)  - ∫dx/(x² + 1)

= x   - ln|x|  + (1/2) ln|x² + 1|  - tan⁻¹x  + C

∫(x³-1)/(x³+x) dx = x   - ln|x|  + (1/2) ln|x² + 1|  - tan⁻¹x  + C

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Answered by khushh213
0

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