Question
\( \int \frac { 3 } { 1 + x ^ { 2 } } d x = [ ? ] [ ] ^ { - 1 } x + C \)
Ask by Weston Guzman. in the United States
Mar 07,2025
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Tutor-Verified Answer
Answer
The integral is \( 3\arctan(x) + C \).
Solution
Calculate the integral \( \int \frac{3}{1+x^2}dx \).
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
\(\int \frac{3}{1+x^{2}} dx\)
- step1: Rewrite the expression:
\(\int 3\times \frac{1}{1+x^{2}} dx\)
- step2: Use properties of integrals:
\(3\times \int \frac{1}{1+x^{2}} dx\)
- step3: Evaluate the integral:
\(3\arctan\left(x\right)\)
- step4: Add the constant of integral C:
\(3\arctan\left(x\right) + C, C \in \mathbb{R}\)
The integral of \( \frac{3}{1+x^2} \) with respect to \( x \) is \( 3\arctan(x) + C \), where \( C \) is the constant of integration.
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Simplify this solution The Deep Dive
To solve the integral \( \int \frac { 3 } { 1 + x ^ { 2 } } d x \), you can recognize that the integral has the form \( \frac{1}{1+x^2} \), which directly relates to the arctangent function. Specifically, \( \int \frac{1}{1+x^2} \, dx = \tan^{-1}(x) + C \). So, upon integrating, you get: \[ \int \frac { 3 } { 1 + x ^ { 2 } } d x = 3 \tan^{-1}(x) + C \] Now you can fill in the blank: \[ 3 \tan^{-1}(x) + C \] So, the solution can be expressed as: \[ \int \frac { 3 } { 1 + x ^ { 2 } } d x = 3 \tan^{-1}(x) + C \] And here, the answer to your question is \(3\) for the first blank and \(\tan\) for the second blank, giving the answer of \(3 [ \tan ] ^{ -1 } x + C\).
