Identify integral: Identify the integral that needs to be solved. We need to find the integral of the function tan^(-1)(5x) with respect to x. ∫tan^(-1)(5x)dxUse integration by parts: Use integration by parts. Integration by parts formula is ∫u dv = uv - ∫v du, where u and dv are parts of the integrand. Let u = tan^(-1)(5x), which means du = (5/(1+(5x)^2))dx. Let dv = dx, which means v = x.Differentiate and integrate: Differentiate u and integrate dv. Differentiate u to find du: du = d/dx[tan^(-1)(5x)]dx = (5/(1+(5x)^2))dx Integrate dv to find v: v = ∫dx = xApply integration by parts: Apply the integration by parts formula. Now we have u, du, v, and dv, we can apply the integration by parts formula: ∫tan^(-1)(5x)dx = uv - ∫v du = (tan^(-1)(5x)) * x - ∫x * (5/(1+(5x)^2))dxSimplify the integral: Simplify the integral. We need to simplify the integral ∫x * (5/(1+(5x)^2))dx. Let's make a substitution to simplify this integral. Let w = 1+(5x)^2, then dw = 10xdx.Change variables: Change the variables in the integral. Substitute w and dw into the integral: ∫x * (5/(1+(5x)^2))dx = 1/2 * ∫(5/(w))dw We have the extra 1/2 because dw = 10xdx, so we need to multiply by 1/10 to get the original xdx, and since we have a 5 in the numerator, 5 * 1/10 = 1/2.Integrate with respect to w: Integrate with respect to w. Now integrate 1/2 * ∫(5/w)dw: 1/2 * ∫(5/w)dw = 1/2 * 5 * ln|w| + C = (5/2) * ln|w| + CSubstitute back for x: Substitute back for x. Now we need to substitute back in for w to get the integral in terms of x: (5/2) * ln|w| + C = (5/2) * ln|1+(5x)^2| + CCombine integration results: Combine the results from integration by parts. Combine the result from Step 4 and Step 8 to get the final answer: ∫tan^(-1)(5x)dx = (tan^(-1)(5x)) * x - (5/2) * ln|1+(5x)^2| + C

Algebra 1

Evaluate integers raised to rational exponents

Full solution

\int \tan ^{-1}(5 x) d x

Identify integral: Identify the integral that needs to be solved. $\newline$ We need to find the integral of the function $\tan^{-1}(5x)$ with respect to $x$ . $\newline$ $\int \tan^{-1}(5x)\,dx$
Use integration by parts: Use integration by parts. $\newline$ Integration by parts formula is $\int u \, dv = uv - \int v \, du$ , where $u$ and $dv$ are parts of the integrand. $\newline$ Let $u = \tan^{-1}(5x)$ , which means $du = \left(\frac{5}{1+(5x)^2}\right)dx$ . $\newline$ Let $dv = dx$ , which means $v = x$ .
Differentiate and integrate: Differentiate $u$ and integrate $dv$ .\ Differentiate $u$ to find $du$ :\ $du = \frac{d}{dx}[\tan^{-1}(5x)]dx = \left(\frac{5}{1+(5x)^2}\right)dx$ \ Integrate $dv$ to find $v$ :\ $v = \int dx = x$
Apply integration by parts: Apply the integration by parts formula. $\newline$ Now we have $u$ , $du$ , $v$ , and $dv$ , we can apply the integration by parts formula: $\newline$ $\int \tan^{-1}(5x)\,dx = uv - \int v\,du$ $\newline$ = $(\tan^{-1}(5x)) \cdot x - \int x \cdot \left(\frac{5}{1+(5x)^2}\right)dx$
Simplify the integral: Simplify the integral. $\newline$ We need to simplify the integral $\int x \cdot \left(\frac{5}{1+(5x)^2}\right)dx$ . $\newline$ Let's make a substitution to simplify this integral. Let $w = 1+(5x)^2$ , then $dw = 10xdx$ .
Change variables: Change the variables in the integral. $\newline$ Substitute $w$ and $dw$ into the integral: $\newline$ $\int x \cdot \left(\frac{5}{1+(5x)^2}\right)dx = \frac{1}{2} \cdot \int\left(\frac{5}{w}\right)dw$ $\newline$ We have the extra $\frac{1}{2}$ because $dw = 10xdx$ , so we need to multiply by $\frac{1}{10}$ to get the original $xdx$ , and since we have a $5$ in the numerator, $5 \cdot \frac{1}{10} = \frac{1}{2}$ .
Integrate with respect to $w$ : Integrate with respect to $w$ . Now integrate $\frac{1}{2} \times \int\left(\frac{5}{w}\right)dw$ : \frac{$1$}{$2$} \times \int\left(\frac{$5$}{w}\right)dw = \frac{$1$}{$2$} \times \(5 \times \ln|w| + C = \left(\frac{ $5$ }{ $2$ }\right) \times \ln|w| + C
Substitute back for x: Substitute back for x. $\newline$ Now we need to substitute back in for $w$ to get the integral in terms of $x$ : $\newline$ $\frac{5}{2} \cdot \ln|w| + C = \frac{5}{2} \cdot \ln|1+(5x)^2| + C$
Combine integration results: Combine the results from integration by parts. $\newline$ Combine the result from Step $4$ and Step $8$ to get the final answer: $\newline$ $\int \tan^{-1}(5x)\,dx = (\tan^{-1}(5x)) \cdot x - \frac{5}{2} \cdot \ln|1+(5x)^2| + C$