Solve ∫ sqrt{y}(sqrt{y}+1/sqrt{y}) wrt y

$$\int \sqrt { y } ( \sqrt { y } + \frac { 1 } { \sqrt { y } } ) d y$$

Evaluate

$$\frac{y^{2}}{2}+y+С$$

Solution Steps

Use the distributive property to multiply \sqrt{y} by \sqrt{y}+\frac{1}{\sqrt{y}}.

$$\int \left(\sqrt{y}\right)^{2}+\sqrt{y}\times \frac{1}{\sqrt{y}}\mathrm{d}y$$

Calculate \sqrt{y} to the power of 2 and get y.

$$\int y+\sqrt{y}\times \frac{1}{\sqrt{y}}\mathrm{d}y$$

Cancel out \sqrt{y} and \sqrt{y}.

$$\int y+1\mathrm{d}y$$

Integrate the sum term by term.

$$\int y\mathrm{d}y+\int 1\mathrm{d}y$$

Since \int y^{k}\mathrm{d}y=\frac{y^{k+1}}{k+1} for k\neq -1, replace \int y\mathrm{d}y with \frac{y^{2}}{2}.

$$\frac{y^{2}}{2}+\int 1\mathrm{d}y$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}y=ay.

$$\frac{y^{2}}{2}+y$$

If F\left(y\right) is an antiderivative of f\left(y\right), then the set of all antiderivatives of f\left(y\right) is given by F\left(y\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\frac{y^{2}}{2}+y+С$$

Differentiate w.r.t. y

$$y+1$$