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Question: How do you integrate \[\dfrac{x}{{{x^2} + 1}}\]...

How do you integrate xx2+1\dfrac{x}{{{x^2} + 1}}

Explanation

Solution

We need to identify the integration of xx2+1\dfrac{x}{{{x^2} + 1}} in this question, therefore we must first understand what integration is. In this question we will use a substitution method to solve the given integral. We integrate a function when we are given a differentiated function and we need to determine the function. Different values of integral may be obtained by changing the value of an arbitrary constant; therefore, a function can have an unlimited number of integrals, but each function has a unique derivative.

Complete step by step answer:
The given function is ∫xx2+1 dx\int {\dfrac{x}{{{x^{^2}} + 1}}} \,dx
Let x2+1=t{x^2} + 1 = t
Differentiating both the sides with respect to xx
ddx(x2+1)=dtdx\dfrac{d}{{dx}}({x^2} + 1) = \dfrac{{dt}}{{dx}}
On differentiating we get
⇒2x dx=dt\Rightarrow 2x\,dx = dt
Now we will re-arrange the differentiated function in accordance with our integral function
⇒x dx=12 dt\Rightarrow x\,dx = \dfrac{1}{2}\,dt
Now we substitute this in place of x dxx\,dx in the given integral function,
I=∫12 dtt I = \int {\dfrac{1}{2}\,\dfrac{{dt}}{t}\,}
We know that integral of logarithmic function is ∫dxx=ln⁡∣x∣+C\int {\dfrac{{dx}}{x} = \ln |x| + C} .
Thus, the above integral function is equal to 12ln⁡∣t∣+C\dfrac{1}{2}\ln |t| + C
Substituting the value of tt back in the equation
Thus, 12ln⁡∣x2+1∣+C\dfrac{1}{2}\ln |{x^2} + 1| + C
Hence the integration of the function xx2+1\dfrac{x}{{{x^2} + 1}} is 12ln⁡∣x2+1∣+C\dfrac{1}{2}\ln |{x^2} + 1| + C.

Note: The function whose integration we must determine in this question is a fraction with polynomials in both the numerator and denominator. In this question, we utilized the substitution method to solve the function. Substitution methods help in solving the problem easily. Differentiation also plays an important role. Since, we are utilizing both derivation and integration formulas in these sorts of problems, students should not be confused about where to use the formulae and which one to utilize because both derivations and integrations include numerous formulas. The following are some of the most crucial formulae to know:
ddxxn=nxn−1\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}
∫dxx=ln⁡∣x∣+C\int {\dfrac{{dx}}{x} = \ln |x| + C}