/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 22

Evaluate the following integrals. $$\int x \tan ^{-1} x^{2} d x$$

Short Answer

Expert verified
Question: Find the integral of the function \(x \tan^{-1}(x^2)\). Answer: The integral of the function \(x \tan^{-1}(x^2)\) is \(\frac{x^2}{2}\tan^{-1}(x^2) - \frac{1}{4}\ln(1+x^4) + C\).

Step by step solution

01

Choose u and dv

Let's choose the following functions: $$u = \tan^{-1}(x^2) \Rightarrow du = \frac{2x}{1 + x^4} dx$$ $$dv = x dx \Rightarrow v = \frac{x^2}{2}$$
02

Apply integration by parts formula

Now, let's apply the integration by parts formula: $$\int x \tan^{-1}(x^2) dx = \int u dv = uv - \int v du$$ Substitute the values of u, du, v, and dv from Step 1: $$ = \frac{x^2}{2}\tan^{-1}(x^2)-\int\frac{x^2}{2}\left(\frac{2x}{1+x^4}\right) dx$$
03

Simplify the integral

Now, let's simplify the integral: $$ = \frac{x^2}{2}\tan^{-1}(x^2)-\int\frac{x^3}{1+x^4} dx$$
04

Substitution

Let's make a substitution: $$t = x^2 \Rightarrow dt = 2x dx \Rightarrow \frac{1}{2} dt = x dx$$ The integral becomes: $$ = \frac{x^2}{2}\tan^{-1}(x^2)-\frac{1}{2}\int\frac{t}{1+t^2} dt$$
05

Integrate

Now, we can integrate the function: $$ = \frac{x^2}{2}\tan^{-1}(x^2)-\frac{1}{2}\left(\int\frac{t}{1+t^2} dt\right)$$ The integral of \(\frac{t}{1+t^2}\) is \(\frac{1}{2}\ln(1+t^2)\). Therefore, $$ = \frac{x^2}{2}\tan^{-1}(x^2)-\frac{1}{2}\left(\frac{1}{2}\ln(1+t^2)\right)$$
06

Undo substitution

Now, let's undo the substitution: $$ = \frac{x^2}{2}\tan^{-1}(x^2) - \frac{1}{4}\ln(1+x^4)$$
07

Add constant of integration

Finally, we add the constant of integration: $$ = \frac{x^2}{2}\tan^{-1}(x^2) - \frac{1}{4}\ln(1+x^4) + C$$ So the integral of \(x \tan^{-1}(x^2)\) is $$\int x \tan ^{-1} x^{2} d x = \frac{x^2}{2}\tan^{-1}(x^2) - \frac{1}{4}\ln(1+x^4) + C$$

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Most popular questions from this chapter

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