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int(1)/(xsqrt(x+1))dx

$\int \frac{1}{x \sqrt{x+1}} d x$

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Find indefinite integrals using the substitution and by parts

Full solution

Q.

\int \frac{1}{x \sqrt{x+1}} d x

Substitution: Let's do a substitution. Let $u = x + 1$ , which means $du = dx$ and $x = u - 1$ .
Split Fraction: Now we substitute $x$ and $dx$ in the integral: $\int \frac{1}{(u - 1)\sqrt{u}} \, du$ .
Simplify Terms: This looks a bit messy, let's split the fraction: $\int(\frac{1}{u\sqrt{u}} - \frac{1}{u^2\sqrt{u}}) \, du$ .
Integrate Term by Term: Simplify the terms: $\int(\frac{1}{u^{\frac{3}{2}}} - \frac{1}{u^{\frac{5}{2}}}) \, du$ .
Integrate First Term: Now we can integrate term by term: $\int \frac{1}{u^{3/2}} \, du - \int \frac{1}{u^{5/2}} \, du$ .
Integrate Second Term: Integrate the first term: $-2/u^{1/2}$ (because $\int u^{-n} du = u^{-n+1}/(-n+1)$ ).
Combine Terms: Integrate the second term: $\frac{2}{3}u^{\frac{3}{2}}$ (because $\int u^{-m} du = \frac{u^{-m+1}}{-m+1}$ ).
Substitute Back: Combine the two terms: $-\frac{2}{u^{1/2}} + \frac{2}{3}u^{3/2} + C$ , where $C$ is the constant of integration.
Correct Integration: Substitute back $u = x + 1$ : $-\frac{2}{(x + 1)^{\frac{1}{2}}} + \frac{2}{3}(x + 1)^{\frac{3}{2}} + C$ .
Correct Integration: Substitute back $u = x + 1$ : $-\frac{2}{(x + 1)^{\frac{1}{2}}} + \frac{2}{3}(x + 1)^{\frac{3}{2}} + C$ .Oops, I made a mistake in the integration step. The correct integration of $\frac{1}{u^{\frac{3}{2}}}$ should be $-\frac{2}{u^{\frac{1}{2}}},$ but for $\frac{1}{u^{\frac{5}{2}}},$ it should be $\frac{2}{3u^{\frac{3}{2}}},$ not $\frac{2}{3}u^{\frac{3}{2}}$ . Let's correct that.

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