QuestionFind the integral of $\frac{\tan^{-1} x}{1 + \frac{1}{x^2}}$ with respect to $x$ .

Studdy Solution

STEP 1

Assumptions1. The integral is a definite integral. . The integral is of the form $\int \frac{f'(x)}{f(x)} dx$ , which can be solved using the logarithmic rule of integration.
3. The function $f(x)$ is $1+\frac{1}{x^{}}$ and its derivative $f'(x)$ is $\tan ^{-1} x$ .

STEP 2

First, we need to check if the derivative of the function $f(x)$ is indeed $\tan ^{-1} x$ . We can do this by differentiating $f(x)$ .
$f'(x) = \frac{d}{dx} \left(1+\frac{1}{x^{2}}\right)$

STEP 3

Now, differentiate the function $f(x)$ .
$f'(x) =0 - \frac{2}{x^3}$

STEP 4

We see that the derivative of $f(x)$ is not $\tan ^{-1} x$ . Therefore, our initial assumption was incorrect. We need to find a new function $f(x)$ and its derivative $f'(x)$ that fit the form of the integral.

STEP 5

Let's rewrite the integral in a more convenient form. Notice that $\tan ^{-1} x$ is the derivative of $1+\frac{1}{x^{2}}$ .
$\int \frac{\tan ^{-1} x}{1+\frac{1}{x^{2}}} dx = \int \frac{d(1+\frac{1}{x^{2}})}{1+\frac{1}{x^{2}}}$

STEP 6

Now, we can use the logarithmic rule of integration, which states that $\int \frac{f'(x)}{f(x)} dx = \ln |f(x)| + C$ .
$\int \frac{d(1+\frac{1}{x^{2}})}{1+\frac{1}{x^{2}}} = \ln |1+\frac{1}{x^{2}}| + C$

STEP 7

The absolute value is unnecessary in this case because $1+\frac{1}{x^{2}}$ is always positive. Therefore, the integral simplifies to $\int \frac{\tan ^{-1} x}{1+\frac{1}{x^{2}}} dx = \ln (1+\frac{1}{x^{2}}) + C$ So, the solution to the integral $\int \frac{\tan ^{-1} x}{1+\frac{1}{x^{2}}} dx$ is $\ln (1+\frac{1}{x^{2}}) + C$ .

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QuestionFind the integral of $\frac{\tan^{-1} x}{1 + \frac{1}{x^2}}$ with respect to $x$ .

Studdy Solution

STEP 1

Assumptions1. The integral is a definite integral. . The integral is of the form $\int \frac{f'(x)}{f(x)} dx$ , which can be solved using the logarithmic rule of integration.
3. The function $f(x)$ is $1+\frac{1}{x^{}}$ and its derivative $f'(x)$ is $\tan ^{-1} x$ .

STEP 2

First, we need to check if the derivative of the function $f(x)$ is indeed $\tan ^{-1} x$ . We can do this by differentiating $f(x)$ .
$f'(x) = \frac{d}{dx} \left(1+\frac{1}{x^{2}}\right)$

STEP 3

Now, differentiate the function $f(x)$ .
$f'(x) =0 - \frac{2}{x^3}$

STEP 4

We see that the derivative of $f(x)$ is not $\tan ^{-1} x$ . Therefore, our initial assumption was incorrect. We need to find a new function $f(x)$ and its derivative $f'(x)$ that fit the form of the integral.

STEP 5

Let's rewrite the integral in a more convenient form. Notice that $\tan ^{-1} x$ is the derivative of $1+\frac{1}{x^{2}}$ .
$\int \frac{\tan ^{-1} x}{1+\frac{1}{x^{2}}} dx = \int \frac{d(1+\frac{1}{x^{2}})}{1+\frac{1}{x^{2}}}$

STEP 6

Now, we can use the logarithmic rule of integration, which states that $\int \frac{f'(x)}{f(x)} dx = \ln |f(x)| + C$ .
$\int \frac{d(1+\frac{1}{x^{2}})}{1+\frac{1}{x^{2}}} = \ln |1+\frac{1}{x^{2}}| + C$

STEP 7

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QuestionFind the integral of tan⁡−1x1+1x2\frac{\tan^{-1} x}{1 + \frac{1}{x^2}}1+x21​tan−1x​ with respect to xxx.

Studdy Solution

STEP 1

STEP 2

First, we need to check if the derivative of the function f(x)f(x)f(x) is indeed tan⁡−1x\tan ^{-1} xtan−1x. We can do this by differentiating f(x)f(x)f(x).f′(x)=ddx(1+1x2)f'(x) = \frac{d}{dx} \left(1+\frac{1}{x^{2}}\right)f′(x)=dxd​(1+x21​)

STEP 3

Now, differentiate the function f(x)f(x)f(x).f′(x)=0−2x3f'(x) =0 - \frac{2}{x^3}f′(x)=0−x32​

STEP 4

We see that the derivative of f(x)f(x)f(x) is not tan⁡−1x\tan ^{-1} xtan−1x. Therefore, our initial assumption was incorrect. We need to find a new function f(x)f(x)f(x) and its derivative f′(x)f'(x)f′(x) that fit the form of the integral.

STEP 5

STEP 6

STEP 7

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QuestionFind the integral of $\frac{\tan^{-1} x}{1 + \frac{1}{x^2}}$ with respect to $x$ .

First, we need to check if the derivative of the function $f(x)$ is indeed $\tan ^{-1} x$ . We can do this by differentiating $f(x)$ .
$f'(x) = \frac{d}{dx} \left(1+\frac{1}{x^{2}}\right)$

Now, differentiate the function $f(x)$ .
$f'(x) =0 - \frac{2}{x^3}$

We see that the derivative of $f(x)$ is not $\tan ^{-1} x$ . Therefore, our initial assumption was incorrect. We need to find a new function $f(x)$ and its derivative $f'(x)$ that fit the form of the integral.