$\int \frac{d x}{3 \sqrt{x-1}-4 \sqrt[3]{x-1}}$

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Find indefinite integrals using the substitution and by parts

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\int \frac{d x}{3 \sqrt{x-1}-4 \sqrt[3]{x-1}}

Write Integral: Write down the integral to be solved. $\newline$ $I = \int \frac{dx}{3\sqrt{x-1} - 4\sqrt[3]{x-1}}$
Make Substitution: To simplify the integral, let's make a substitution. Let $u = x - 1$ , which implies that $du = dx$ . $\newline$ $I = \int \frac{du}{3\sqrt{u} - 4\sqrt[3]{u}}$
Rewrite in Terms of u: Rewrite the integral in terms of u. $\newline$ $I = \int \frac{du}{3u^{\frac{1}{2}} - 4u^{\frac{1}{3}}}$
Simplify Denominator: To integrate this expression, we need to find a way to simplify the denominator. One common technique is to multiply the numerator and the denominator by the conjugate of the denominator to rationalize it. However, in this case, the conjugate involves complex expressions, and it might not lead to a simpler form. Instead, we should look for a different approach, such as partial fraction decomposition or another substitution.
Try Another Substitution: Since the denominator has terms with $u$ raised to fractional powers, partial fraction decomposition is not straightforward. We need to consider another substitution or method to simplify the integral. Let's try another substitution to simplify the denominator. Let's denote $v = u^{\frac{1}{6}}$ , which implies $u = v^6$ and $du = 6v^5 dv$ . $\newline$ $I = \int \frac{6v^5 dv}{3v^3 - 4v^2}$
Cancel Common Factors: Simplify the integral by canceling out common factors. $\newline$ $I = \int\frac{6v^5 \, dv}{3v^2(v - \frac{4}{3})}$ $\newline$ $I = 2\int\frac{v^3 \, dv}{v - \frac{4}{3}}$
Perform Polynomial Division: Now, we can perform long division or use polynomial division to simplify the integral further. Let's divide $v^3$ by $v - \frac{4}{3}$ .
Integrate Each Term: Perform the polynomial division. $\newline$ $v^3$ divided by $v$ gives $v^2$ . $\newline$ Multiply $v^2$ by $(v - \frac{4}{3})$ to get $v^3 - \frac{4}{3}v^2$ . $\newline$ Subtract this from $v^3$ to get $\frac{4}{3}v^2$ . $\newline$ Now, divide $\frac{4}{3}v^2$ by $v$ to get $v$ $0$ . $\newline$ Multiply $v$ $0$ by $(v - \frac{4}{3})$ to get $v$ $3$ . $\newline$ Subtract this from $\frac{4}{3}v^2$ to get $v$ $5$ . $\newline$ Finally, divide $v$ $5$ by $v$ to get $v$ $8$ . $\newline$ Multiply $v$ $8$ by $(v - \frac{4}{3})$ to get $v^2$ $1$ . $\newline$ Subtract this from $v$ $5$ to get a remainder of $v^2$ $3$ . $\newline$ So, $v^3$ divided by $(v - \frac{4}{3})$ is $v^2$ $6$ with a remainder of $v^2$ $3$ .
Substitute Back to x: Rewrite the integral with the result of the polynomial division. I = 2\int(v^2 + \frac{4}{3}v + \frac{16}{9}) dv - 2\int\frac{64}{27}}{v - \frac{4}{3}} dv
Substitute Back to x: Rewrite the integral with the result of the polynomial division. $\newline$ $I = 2\int(v^2 + \frac{4}{3}v + \frac{16}{9}) dv - 2\int(\frac{64}{27})/(v - \frac{4}{3}) dv$ Integrate each term separately. $\newline$ $I = 2[\frac{v^3}{3} + (\frac{4}{3})(\frac{v^2}{2}) + (\frac{16}{9})v] - 2*(\frac{64}{27})\ln|v - \frac{4}{3}| + C$ $\newline$ $I = 2[\frac{v^3}{3} + (\frac{2}{3})v^2 + (\frac{16}{9})v] - (\frac{128}{27})\ln|v - \frac{4}{3}| + C$
Substitute Back to x: Rewrite the integral with the result of the polynomial division. $\newline$ $I = 2\int(v^2 + \frac{4}{3}v + \frac{16}{9}) dv - 2\int(\frac{64}{27})/(v - \frac{4}{3}) dv$ Integrate each term separately. $\newline$ $I = 2[\frac{(v^3)}{3} + (\frac{4}{3})(\frac{v^2}{2}) + (\frac{16}{9})v] - 2*(\frac{64}{27})\ln|v - \frac{4}{3}| + C$ $\newline$ $I = 2[\frac{(v^3)}{3} + (\frac{2}{3})v^2 + (\frac{16}{9})v] - (\frac{128}{27})\ln|v - \frac{4}{3}| + C$ Substitute back the original variable $u = v^6$ to express the integral in terms of x. $\newline$ $I = 2[\frac{((u^{1/6})^3)}{3} + (\frac{2}{3})(u^{1/6})^2 + (\frac{16}{9})u^{1/6}] - (\frac{128}{27})\ln|u^{1/6} - \frac{4}{3}| + C$ $\newline$ $I = 2[\frac{(u^{1/2})}{3} + (\frac{2}{3})u^{1/3} + (\frac{16}{9})u^{1/6}] - (\frac{128}{27})\ln|u^{1/6} - \frac{4}{3}| + C$
Substitute Back to x: Rewrite the integral with the result of the polynomial division. $\newline$ $I = 2\int(v^2 + \frac{4}{3}v + \frac{16}{9}) dv - 2\int(\frac{64}{27})/(v - \frac{4}{3}) dv$ Integrate each term separately. $\newline$ $I = 2[\frac{(v^3)}{3} + (\frac{4}{3})(\frac{v^2}{2}) + (\frac{16}{9})v] - 2*(\frac{64}{27})\ln|v - \frac{4}{3}| + C$ $\newline$ $I = 2[\frac{(v^3)}{3} + (\frac{2}{3})v^2 + (\frac{16}{9})v] - (\frac{128}{27})\ln|v - \frac{4}{3}| + C$ Substitute back the original variable $u = v^6$ to express the integral in terms of x. $\newline$ $I = 2[\frac{((u^{1/6})^3)}{3} + (\frac{2}{3})(u^{1/6})^2 + (\frac{16}{9})u^{1/6}] - (\frac{128}{27})\ln|u^{1/6} - \frac{4}{3}| + C$ $\newline$ $I = 2[\frac{(u^{1/2})}{3} + (\frac{2}{3})u^{1/3} + (\frac{16}{9})u^{1/6}] - (\frac{128}{27})\ln|u^{1/6} - \frac{4}{3}| + C$ Substitute back the original variable $x = u + 1$ to express the integral in terms of x. $\newline$ $I = 2[\frac{((x-1)^{1/2})}{3} + (\frac{2}{3})(x-1)^{1/3} + (\frac{16}{9})(x-1)^{1/6}] - (\frac{128}{27})\ln|(x-1)^{1/6} - \frac{4}{3}| + C$

∫dx3x−1−4x−13 \int \frac{d x}{3 \sqrt{x-1}-4 \sqrt[3]{x-1}} ∫3x−1​−43x−1​dx​

$\int \frac{d x}{3 \sqrt{x-1}-4 \sqrt[3]{x-1}}$