Integration by Substitution #engineering #maths #college

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Rewriting integral:
∫dx/(x³√(1+9/x²))
Substitute u = √(1+9/x²) ----(i)
du = -9/(x³√(1+9/x²)) dx
-du/9 = dx/(x³√(1+9/x²))
Thus the integral becomes:
-1/9 ∫du
= -u/9
Substitute from (i)
(-1/9)√(1+9/x²)+c
= (-1/9x)√(9+x²)
Thus, -√(9+x²)/9x + C

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