Solve ∫ frac{sqrt{x}}{sqrt{b+1 wrt x=

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

Evaluate

$$\frac{2x^{\frac{3}{2}}}{3\sqrt{b+1}}+С$$

Solution Steps

Factor out the constant using \int af\left(x\right)\mathrm{d}x=a\int f\left(x\right)\mathrm{d}x.

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

Rewrite \sqrt{x} as x^{\frac{1}{2}}. Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{\frac{1}{2}}\mathrm{d}x with \frac{x^{\frac{3}{2}}}{\frac{3}{2}}. Simplify.

$$\frac{2x^{\frac{3}{2}}}{3\sqrt{b+1}}$$

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

$$\frac{2x^{\frac{3}{2}}}{3\sqrt{b+1}}+С$$

Differentiate w.r.t. x

$$\sqrt{\frac{x}{b+1}}$$