Question

Use the indicated substitution to convert the given integral to an integral of a rational function. Evaluate the resulting integral. $$\int \frac{d x}{\sqrt[4]{x+2}+1} ; x+2=u^{4}$$

Short answer

Question: Find the indefinite integral: $$\int{\frac{dx}{\sqrt[4]{x+2}+1}},$$ given that \(x+2=u^4\). Answer: $$F(x) = -4\left[\frac{1}{3}(\sqrt[4]{x+2})^3- (\sqrt[4]{x+2})^2+ 3\sqrt[4]{x+2}\right]-2(\sqrt[4]{x+2})^2+c$$

Step-by-step solution

Perform the substitution and change the integral bounds

Since it's given that \(x+2=u^4\), we can write \(x=u^4-2\). Now, we can compute the derivative of \(x\) with respect to \(u\). By taking the derivative, we get \(\frac{dx}{du}=4u^3\). So, \({\frac{d x}{d u}} d u={\frac{d x}{d u}}(du)={dx}\). Now, substitute the values of \(x\) and \(dx\) in the given integral: $$\int{\frac{dx}{\sqrt[4]{x+2}+1}}= \int{\frac{4u^3 du}{\sqrt[4]{u^4}+1}}$$

Simplify the integral and find the integral of the rational function.

The integral can be simplified as follows: $$\int{\frac{4u^3 du}{\sqrt[4]{u^4}+1}}= \int{\frac{4u^3 du}{u+1}}$$ Now we have a rational function, and we can integrate it using partial fractions: $$\int\frac{4u^3}{u+1}du=-4\int u^2-2u+3-2du$$ Now, we can integrate term by term: $$\: =-4[u^3/3- u^2 + 3u]-2(u^2/2+c)$$

Perform the inverse substitution and find the final answer

The final step is to change back the variable to \(x\). Recall that \(u^{4}=x+2,\) so \(u=\sqrt[4]{x+2}\). Substituting back, we get: $$ -4\left[\frac{1}{3}(\sqrt[4]{x+2})^3- (\sqrt[4]{x+2})^2+ 3\sqrt[4]{x+2}\right]-2(\sqrt[4]{x+2})^2+c$$ The final answer to the given integral is: $$F(x) = -4\left[\frac{1}{3}(\sqrt[4]{x+2})^3- (\sqrt[4]{x+2})^2+ 3\sqrt[4]{x+2}\right]-2(\sqrt[4]{x+2})^2+c$$