$\int_{0}^{1} \frac{x^{3}}{1+x^{2}} d x=$

Question

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To integrate the function $$\int_{0}^{1} \frac{x^{3}}{1+x^{2}} dx$$, we look for a method to simplify the integrand. ‖ ‖ step 2 ⋮ We notice that the derivative of $$1+x^2$$ is $$2x$$, which is close to the numerator $$x^3$$. This suggests a substitution. ‖ ‖ step 3 ⋮ Let $$u = 1+x^2$$. Then, $$du = 2x dx$$. ‖ ‖ step 4 ⋮ We need to adjust the numerator to match our substitution. We can write $$x^3 dx$$ as $$\frac{1}{2}x^2 \cdot 2x dx$$. ‖ ‖ step 5 ⋮ Substituting $$u$$ and $$du$$ into the integral, we get $$\frac{1}{2}\int \frac{u-1}{u} du$$. ‖ ‖ step 6 ⋮ Split the integral: $$\frac{1}{2}\int 1 - \frac{1}{u} du$$. ‖ ‖ step 7 ⋮ Integrate term by term: $$\frac{1}{2}(u - \ln|u|) + C$$. ‖ ‖ step 8 ⋮ Substitute back for $$u$$: $$\frac{1}{2}(x^2 - \ln|x^2+1|) + C$$. ‖ ‖ step 9 ⋮ Evaluate the definite integral from 0 to 1: $$\frac{1}{2}[(1^2 - \ln|1^2+1|) - (0^2 - \ln|0^2+1|)]$$. ‖ ‖ step 10 ⋮ Simplify to get the final answer: $$\frac{1}{2}(1 - \ln(2))$$. ‖ ∻Answer∻ ⚹ $$\frac{1}{2}(1 - \ln(2)) \approx 0.15343$$ ⚹ ∻Key Concept∻ ⚹Integration by Substitution⚹ ∻Explanation∻ ⚹To integrate a function of the form $$\frac{x^n}{1+x^2}$$, where $$n$$ is an odd integer, we can use substitution with $$u = 1+x^2$$, simplifying the integration process.⚹◊What is the indefinite integral of $\frac{x^{3}}{1+x^{2}}$?⍭ Generate me a similar question◊

Sia