(i) tat int(xtan^(-1)x)/((1+x^(2))^(3//2))dx

Identify integral: Identify the integral to be solved. We need to evaluate the integral of the function (xtan^(-1)x)/((1+x^(2))^(3/2)) with respect to x.Choose substitution: Choose a substitution. Let u = tan^(-1)x, which implies x = tan(u). Then, we need to find dx in terms of du.Differentiate u: Differentiate u with respect to x to find du/dx. du/dx = 1/(1+x^2) dx = (1+x^2)duSubstitute x and dx: Substitute x and dx in the integral. The integral becomes: ∫(tan(u) * u) / ((1+tan^2(u))^(3/2)) * (1+tan^2(u))duSimplify integral: Simplify the integral. Since 1+tan^2(u) = sec^2(u), the integral simplifies to: ∫(tan(u) * u) / (sec^3(u)) * sec^2(u)du ∫(tan(u) * u) / sec(u)du ∫(sin(u) * u) / cos^2(u)duPerform integration by parts: Perform integration by parts. Let v = u and dw = sin(u)/cos^2(u)du. Then we need to find dv and w. dv = du w = ∫sin(u)/cos^2(u)duFind w: Find w by integrating dw. w = ∫sin(u)/cos^2(u)du w = ∫sec(u)tan(u)du w = sec(u) + CApply integration by parts: Apply integration by parts using the formula ∫udv = uv - ∫vdu. ∫(sin(u) * u) / cos^2(u)du = uv - ∫vdu = u * sec(u) - ∫sec(u)du = u * sec(u) - ln|sec(u) + tan(u)| + CSubstitute back x: Substitute back the original variable x. u = tan^(-1)x, sec(u) = sqrt(1+x^2), tan(u) = x The integral becomes: tan^(-1)x * sqrt(1+x^2) - ln|sqrt(1+x^2) + x| + C

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$\int \frac{x\tan^{-1}x}{(1+x^{2})^{\frac{3}{2}}}dx$

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Calculus

Evaluate definite integrals using the chain rule

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\int \frac{x\tan^{-1}x}{(1+x^{2})^{\frac{3}{2}}}dx

Identify integral: Identify the integral to be solved. $\newline$ We need to evaluate the integral of the function $\frac{x\tan^{-1}x}{(1+x^{2})^{\frac{3}{2}}}$ with respect to $x$ .
Choose substitution: Choose a substitution. $\newline$ Let $u = \tan^{-1}x$ , which implies $x = \tan(u)$ . Then, we need to find $dx$ in terms of $du$ .
Differentiate $u$ : Differentiate $u$ with respect to $x$ to find $\frac{du}{dx}$ . $\frac{du}{dx} = \frac{1}{1+x^2}$ $dx = (1+x^2)du$
Substitute $x$ and $dx$ : Substitute $x$ and $dx$ in the integral. $\newline$ The integral becomes: $\newline$ $\int(\tan(u) \cdot u) / ((1+\tan^2(u))^{3/2}) \cdot (1+\tan^2(u))\,du$
Simplify integral: Simplify the integral. $\newline$ Since $1+\tan^2(u) = \sec^2(u)$ , the integral simplifies to: $\newline$ $\int(\tan(u) \cdot u) / (\sec^3(u)) \cdot \sec^2(u)du$ $\newline$ $\int(\tan(u) \cdot u) / \sec(u)du$ $\newline$ $\int(\sin(u) \cdot u) / \cos^2(u)du$
Perform integration by parts: Perform integration by parts. $\newline$ Let $v = u$ and $dw = \frac{\sin(u)}{\cos^2(u)}du$ . Then we need to find $dv$ and $w$ . $\newline$ $dv = du$ $\newline$ $w = \int \frac{\sin(u)}{\cos^2(u)}du$
Find $w$ : Find $w$ by integrating $dw$ . $w = \int \frac{\sin(u)}{\cos^2(u)}du$ $w = \int \sec(u)\tan(u)du$ $w = \sec(u) + C$
Apply integration by parts: Apply integration by parts using the formula $\int u\,dv = uv - \int v\,du$ . $\int\frac{\sin(u) \cdot u}{\cos^2(u)}\,du = uv - \int v\,du$ = u \cdot \sec(u) - \int \sec(u)\,du= u \cdot \sec(u) - \ln|\sec(u) + \tan(u)| + C
Substitute back $x$ : Substitute back the original variable $x$ .
$u = \tan^{-1}x$ , $\sec(u) = \sqrt{1+x^2}$ , $\tan(u) = x$
The integral becomes:
$\tan^{-1}x \cdot \sqrt{1+x^2} - \ln|\sqrt{1+x^2} + x| + C$