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Math  /  Calculus

Question(d) ∫1eddx[xln⁡x1+x2]dx\int_{1}^{e} \frac{d}{d x}\left[\frac{x \ln x}{1+x^{2}}\right] d x

Studdy Solution

STEP 1

What is this asking? We need to calculate a definite integral of a derivative. Watch out! Don't forget the Fundamental Theorem of Calculus!

STEP 2

1. Apply the Fundamental Theorem of Calculus
2. Evaluate the endpoints

STEP 3

The Fundamental Theorem of Calculus tells us that the integral of a derivative is just the original function evaluated at the bounds of integration.
It's like magic!
We're asked to calculate ∫1eddx[xln⁡x1+x2]dx. \int_{1}^{e} \frac{d}{dx} \left[ \frac{x \ln x}{1+x^2} \right] dx. The Fundamental Theorem of Calculus says ∫abddxf(x) dx=f(b)−f(a). \int_a^b \frac{d}{dx} f(x) \, dx = f(b) - f(a). In our case, f(x)=xln⁡x1+x2f(x) = \frac{x \ln x}{1+x^2}, a=1a = 1, and b=eb = e.
So we **need to evaluate** f(x)f(x) at x=ex = e and x=1x = 1.

STEP 4

Let's **evaluate** f(e)f(e): f(e)=eln⁡e1+e2. f(e) = \frac{e \ln e}{1 + e^2}. Since ln⁡e=1\ln e = 1, we get f(e)=e⋅11+e2=e1+e2. f(e) = \frac{e \cdot 1}{1 + e^2} = \frac{e}{1 + e^2}.

STEP 5

Now let's **evaluate** f(1)f(1): f(1)=1⋅ln⁡11+12=1⋅01+1=02=0. f(1) = \frac{1 \cdot \ln 1}{1 + 1^2} = \frac{1 \cdot 0}{1 + 1} = \frac{0}{2} = 0.

STEP 6

So, our **final integral** is: f(e)−f(1)=e1+e2−0=e1+e2. f(e) - f(1) = \frac{e}{1 + e^2} - 0 = \frac{e}{1 + e^2}.

STEP 7

The value of the definite integral is e1+e2\frac{e}{1 + e^2}.

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