What is the integral of ((sqrt(x)+1)^2)/(x^3) ?

The integral of ((sqrt(x)+1)^2)/(x^3) is -1/x-4/(3x^{3/2)}-1/(2x^2)+C

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integral of ((sqrt(x)+1)^2)/(x^3)

\int \frac{( x + 1 ) ^{2}}{x ^{3}} d x

- \frac{1}{x} - \frac{4}{3 x ^{\frac{3}{2}}} - \frac{1}{2 x ^{2}} + C

Solution steps

\int \frac{( x + 1 ) ^{2}}{x ^{3}} d x

Expand

\frac{( x + 1 ) ^{2}}{x ^{3}} : \frac{1}{x ^{2}} + \frac{2}{x ^{\frac{5}{2}}} + \frac{1}{x ^{3}}

= \int \frac{1}{x ^{2}} + \frac{2}{x ^{\frac{5}{2}}} + \frac{1}{x ^{3}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{x ^{2}} d x + \int \frac{2}{x ^{\frac{5}{2}}} d x + \int \frac{1}{x ^{3}} d x

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

\int \frac{2}{x ^{\frac{5}{2}}} d x = - \frac{4}{3 x ^{\frac{3}{2}}}

\int \frac{1}{x ^{3}} d x = - \frac{1}{2 x ^{2}}

= - \frac{1}{x} - \frac{4}{3 x ^{\frac{3}{2}}} - \frac{1}{2 x ^{2}}

Add a constant to the solution

= - \frac{1}{x} - \frac{4}{3 x ^{\frac{3}{2}}} - \frac{1}{2 x ^{2}} + C

What is the integral of ((sqrt(x)+1)^2)/(x^3) ?
The integral of ((sqrt(x)+1)^2)/(x^3) is -1/x-4/(3x^{3/2)}-1/(2x^2)+C