Use a substitution to change the integral into one that can be found in an integral table. Then evaluate the integral. \( \int \frac{3 x^{3}+3 x+11}{\left(x^{2}+1\right)^{2}} d x \)
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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: \[ \frac{3}{2} \ln(x^2 + 1) + 4 \tan^{-1}(x) + C \]