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Use a substitution to change the integral into one that can be found in an integral table. Then evaluate the integral. \( \int \frac{3 x^{3}+3 x+11}{\left(x^{2}+1\right)^{2}} d x \)

Ask by Lawson Mccarthy. in the United States
Jan 24,2025

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Answer

The integral evaluates to: \[ \frac{3}{2} \ln(x^{2} + 1) + \frac{11x}{2(x^{2} + 1)} + \frac{11}{2} \arctan(x) + C \] where \( C \) is the constant of integration.

Solution

To evaluate the integral \[ \int \frac{3x^{3} + 3x + 11}{(x^{2} + 1)^{2}} \, dx, \] we can use substitution and break it down into simpler parts. ### Step 1: Substitution Let \( u = x^{2} + 1 \). Then, the derivative of \( u \) with respect to \( x \) is: \[ \frac{du}{dx} = 2x \quad \Rightarrow \quad du = 2x \, dx \quad \Rightarrow \quad x \, dx = \frac{du}{2}. \] ### Step 2: Rewrite the Integral Express the numerator \( 3x^{3} + 3x + 11 \) in terms of \( u \): \[ 3x^{3} = 3x(x^{2}) = 3x(u - 1). \] Now, the integral becomes: \[ \int \frac{3x(u - 1) + 3x + 11}{u^{2}} \, dx = \int \frac{3xu - 3x + 3x + 11}{u^{2}} \, dx = \int \left( \frac{3x}{u} + \frac{11}{u^{2}} \right) \, dx. \] ### Step 3: Split the Integral Separate the integral into two parts: \[ \int \frac{3x}{u} \, dx + \int \frac{11}{u^{2}} \, dx. \] ### Step 4: Evaluate Each Part 1. **First Integral:** \[ \int \frac{3x}{u} \, dx = \frac{3}{2} \int \frac{du}{u} = \frac{3}{2} \ln|u| + C_1 = \frac{3}{2} \ln(x^{2} + 1) + C_1. \] 2. **Second Integral:** \[ \int \frac{11}{u^{2}} \, dx = 11 \int \frac{1}{(x^{2} + 1)^{2}} \, dx. \] A standard integral formula gives: \[ \int \frac{1}{(x^{2} + 1)^{2}} \, dx = \frac{x}{2(x^{2} + 1)} + \frac{1}{2} \arctan(x) + C_2. \] Therefore: \[ 11 \left( \frac{x}{2(x^{2} + 1)} + \frac{1}{2} \arctan(x) \right) + C_2. \] ### Step 5: Combine the Results Putting it all together: \[ \int \frac{3x^{3} + 3x + 11}{(x^{2} + 1)^{2}} \, dx = \frac{3}{2} \ln(x^{2} + 1) + \frac{11x}{2(x^{2} + 1)} + \frac{11}{2} \arctan(x) + C, \] where \( C = C_1 + C_2 \) is the constant of integration. **Final Answer:** \[ \boxed{\,\frac{3}{2}\ln(x^{2}+1)\;+\;\frac{11\,x}{2(x^{2}+1)}\;+\;\frac{11}{2}\tan^{-1}(x)\,+\,C\,} \]

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To evaluate the integral \( \int \frac{3 x^{3}+3 x+11}{\left(x^{2}+1\right)^{2}} d x \), we can simplify the expression using substitution. Let's break down the integrand. Notice that \( x^2 + 1 \) suggests a trigonometric substitution, but we can also rearrange the numerator. We can express \( 3x^3 + 3x + 11 \) as: \[ 3x^3 + 3x + 11 = 3\left(x^2 + 1\right)x + 8 \] This allows us to separate the integral into two parts: \[ \int \frac{3\left(x^2 + 1\right)x}{\left(x^2 + 1\right)^{2}} dx + \int \frac{8}{\left(x^{2}+1\right)^{2}} dx \] This simplifies to: \[ 3\int \frac{x}{x^2 + 1} dx + 8\int \frac{1}{(x^2 + 1)^2} dx \] Now, with the first integral, we can use the substitution \( u = x^2 + 1 \) (therefore, \( du = 2x \, dx \)), which results in \[ \frac{3}{2} \int \frac{1}{u} du = \frac{3}{2} \ln |u| + C = \frac{3}{2} \ln |x^2 + 1| + C \] For the second integral, we know \[ \int \frac{1}{(x^2 + 1)^2} dx = \frac{1}{2} \tan^{-1}(x) + C \] Thus, the second part contributes \( 8 \cdot \frac{1}{2} \tan^{-1}(x) = 4 \tan^{-1}(x) \). Combining these results, we end up with: \[ \int \frac{3 x^{3}+3 x+11}{\left(x^{2}+1\right)^{2}} d x = \frac{3}{2} \ln(x^2 + 1) + 4 \tan^{-1}(x) + C \] So the final answer is: \[ \frac{3}{2} \ln(x^2 + 1) + 4 \tan^{-1}(x) + C \]

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