Question

integration of (x+1)^4/x²

Answer 1

Answer :

We know that,

(x + 1)⁴ = x⁴ + 4x³ + 6x² + 4x + 1

So, (x + 1)⁴ / x²

= (x⁴ + 4x³ + 6x² + 4x + 1)/x²

= x² + 4x + 6 + 4/x + 1/x²

So, ∫ (x + 1)⁴/x² dx

= ∫ (x² + 4x + 6 + 4/x + 1/x²) dx

= ∫ x² dx + 4 ∫ x dx + 6 ∫ dx + 4 ∫ dx/x + ∫ dx/x²

= x³/3 + 2x² + 6x + 4 logx - 1/x + c,

where c is integral constant

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