Integrate. $\newline$ $\int \sqrt{x} \cdot \operatorname{arctan} \sqrt{x} d x$

View step-by-step help

Home

Math Problems

Algebra 1

Multiply radical expressions

Full solution

Q. Integrate.

\newline

\int \sqrt{x} \cdot \operatorname{arctan} \sqrt{x} d x

Recognize the integral: Recognize the integral we need to solve. $\newline$ We have the integral of the square root of $x$ times the arc tangent of the square root of $x$ , which is written as: $\newline$ $\int \sqrt{x} \cdot \arctan(\sqrt{x}) \, dx$
Use integration by parts: Use integration by parts. $\newline$ Integration by parts is given by the formula: $\newline$ $\int u \, dv = uv - \int v \, du$ $\newline$ We need to choose $u$ and $dv$ such that $du$ and $v$ are easier to integrate. Let's choose: $\newline$ $u = \arctan(\sqrt{x})$ (since its derivative is simpler) $\newline$ $dv = \sqrt{x} \, dx$ (since its antiderivative is straightforward)
Compute du and v: Compute du and v. $\newline$ To find du, we differentiate u with respect to x: $\newline$ $du = \frac{d}{dx} [\arctan(\sqrt{x})] dx$ $\newline$ Using the chain rule and the derivative of arctan(x), which is $\frac{1}{1+x^2}$ , we get: $\newline$ $du = \frac{1}{2} \cdot \frac{1}{1+x} \cdot \frac{1}{\sqrt{x}} dx$ $\newline$ Now, to find v, we integrate dv: $\newline$ $v = \int \sqrt{x} dx$ $\newline$ $v = \frac{2}{3} \cdot x^{\frac{3}{2}}$
Apply integration by parts formula: Apply the integration by parts formula. $\newline$ Now we have all the parts needed to apply the integration by parts formula: $\newline$ $\int \sqrt{x} \cdot \arctan(\sqrt{x}) \, dx = uv - \int v \, du$ $\newline$ = $\arctan(\sqrt{x}) \cdot \frac{2}{3} \cdot x^{\frac{3}{2}} - \int \frac{2}{3} \cdot x^{\frac{3}{2}} \cdot \frac{1}{2} \cdot \frac{1}{(1+x)} \cdot \frac{1}{\sqrt{x}} \, dx$
Simplify the integral: Simplify the integral. $\newline$ Simplify the expression inside the integral: $\newline$ $\int (\frac{2}{3}) \cdot x^{\frac{3}{2}} \cdot (\frac{1}{2}) \cdot (\frac{1}{1+x}) \cdot (\frac{1}{\sqrt{x}}) \, dx$ $\newline$ = $\int (\frac{1}{3}) \cdot x \cdot (\frac{1}{1+x}) \, dx$ $\newline$ Now we have a simpler integral to solve.
Solve the simplified integral: Solve the simplified integral. $\newline$ To solve $\int \frac{1}{3} \cdot x \cdot \frac{1}{1+x} \, dx$ , we can use a simple substitution: $\newline$ Let $t = 1 + x$ , then $dt = dx$ and $x = t - 1$ . $\newline$ Our integral becomes: $\newline$ $\int \frac{1}{3} \cdot (t - 1) \cdot \frac{1}{t} \, dt$ $\newline$ $= \frac{1}{3} \cdot \int (1 - \frac{1}{t}) \, dt$ $\newline$ $= \frac{1}{3} \cdot (t - \ln|t|)$ $\newline$ Now we substitute back for x: $\newline$ $= \frac{1}{3} \cdot (1 + x - \ln|1 + x|)$
Write the final answer: Write the final answer. $\newline$ Combining the result from integration by parts with the solved integral, we get: $\newline$ $\int \sqrt{x} \cdot \arctan(\sqrt{x}) \, dx = \arctan(\sqrt{x}) \cdot \left(\frac{2}{3}\right) \cdot x^{\frac{3}{2}} - \left(\frac{1}{3}\right) \cdot (1 + x - \ln|1 + x|) + C$ $\newline$ where $C$ is the constant of integration.