# Integrate $\frac{\sqrt {x} }{\sqrt{2x +3}}$

My attempt to the solution: Substitute $$u:=\sqrt{x}$$. Then $$x= u^2$$ and differentiating we get $$2u \ du= dx$$

$$\int \frac{2u^2}{\sqrt{2u^2+3}}du=\int \frac{2u^2+3-3}{\sqrt{2u^2+3}}du=\int \frac{2u^2+3}{\sqrt{2u^+3}}du-\int\frac{3}{\sqrt{2u^2+3}}du$$

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I'm assuming you want to find:

$$\int\sqrt{\frac{x}{2x+3}} \,\mbox{d}x$$

Hint

$$t^2 = \frac{x}{2x+3} \implies x = \frac{3t^2}{1-2t^2}$$

From here, you find $$\mbox{d}x = \ldots \mbox{d}t$$ and you get a rational function to integrate.

• Thank you very much – Nayonika Vats Jun 12 '19 at 17:37

Rather try to make the denominator free of radical sign by substituting,

$$2x+3 = u^2$$

$$x = \frac{u^2-3}{2}$$

$$dx= udu$$

$$I = \int\sqrt{\frac{u^2-3}{2u^2}}udu = \frac{1}{\sqrt2}\int\sqrt{u^2-3} \ du$$

which can be evaluated by using the identity $$\color{blue}{\int\sqrt{u^2-a^2}du = \frac{u}{2}\sqrt{u^2-a^2}-\frac{a^2}{2}\ln|u+\sqrt{u^2-a^2}|+c}$$

• Thank you very much – Nayonika Vats Jun 12 '19 at 17:36
• You're welcome :) – Ak. Jun 12 '19 at 17:37