Home › ACSEE (Form 6) › Advanced Mathematics › Integration › $\displaystyle\int\frac{x}{\sqrt{x+1}}\,dx$ equa…
$\displaystyle\int\frac{x}{\sqrt{x+1}}\,dx$ equals:
A$\frac{2}{3}(x+1)^{3/2}+2\sqrt{x+1}+c$
B$\frac{2}{3}(x+1)^{3/2}-2\sqrt{x+1}+c$
C$\frac{3}{2}(x+1)^{3/2}-2\sqrt{x+1}+c$
D$\frac{2}{3}(x+1)^{1/2}-2\sqrt{x+1}+c$
Answer & Solution
Correct answer: B. $\frac{2}{3}(x+1)^{3/2}-2\sqrt{x+1}+c$
1. Let $u=\sqrt{x+1}$, so $u^2=x+1$, $x=u^2-1$ and $dx=2u\,du$.
2. The integral becomes $\displaystyle\int\frac{u^2-1}{u}\cdot 2u\,du=\int 2(u^2-1)\,du$.
3. Integrating: $\dfrac{2u^3}{3}-2u+c$.
4. Substituting back $u=\sqrt{x+1}$ gives $\dfrac{2}{3}(x+1)^{3/2}-2\sqrt{x+1}+c$.
5. The sign on the second term is minus because $-1$ was integrated; $\sin^{-1}x$ comes from misreading the denominator as $\sqrt{1-x^2}$.
_Source: NECTA ACSEE 2023 Advanced Mathematics 142/1, Question 9: Integration_
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