Question-201184

Question Number 201184 by Calculusboy last updated on 01/Dec/23

Answered by Sutrisno last updated on 01/Dec/23

m i s a l : \sqrt{2 x} + 4 = u \to d x = \sqrt{2 x} d u

= \int \frac{\sqrt{2 x}}{u} . \sqrt{2 x} d u

= \int \frac{{(u - 4)}^{2}}{u} d u

= \int \frac{u^{2} - 8 u + 16}{u} d u

= \int u - 8 + \frac{16}{u} d u

= \frac{1}{2} u^{2} - 8 u + 16 l n u + c

= \frac{1}{2} {(\sqrt{2 x} + 4)}^{2} - 8 (\sqrt{2 x} + 4) + 16 l n (\sqrt{2 x} + 4) + c

Commented by Calculusboy last updated on 01/Dec/23

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