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F(x)=int_(1)^(2x)(t^(2))/(t^(2)+1)dt F^(')(x)=

F(x)=∫12xt2t2+1dt F(x)=\int_{1}^{2 x} \frac{t^{2}}{t^{2}+1} d t \newlineF′(x)= F^{\prime}(x)=

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Evaluate definite integrals using the power ruleright chevron icon

Full solution

Q. F(x)=∫12xt2t2+1dt F(x)=\int_{1}^{2 x} \frac{t^{2}}{t^{2}+1} d t \newlineF′(x)= F^{\prime}(x)=
  1. Apply Fundamental Theorem of Calculus: Use the Fundamental Theorem of Calculus Part 11 which states that if F(x)=∫axf(t) dtF(x) = \int_{a}^{x}f(t)\,dt, then F′(x)=f(x)F'(x) = f(x). However, since the upper limit of the integral is 2x2x, we need to apply the chain rule.
  2. Differentiate Upper Limit: Differentiate the upper limit of the integral with respect to xx, which is 2x2x. The derivative is 22.
  3. Apply Chain Rule: Now apply the chain rule: F′(x)=f(2x)⋅d(2x)dxF'(x) = f(2x) \cdot \frac{d(2x)}{dx}. Substitute t=2xt = 2x into the integrand and multiply by the derivative of 2x2x.
  4. Substitute and Multiply: F′(x)=(2x)22x2+1×2F'(x) = \frac{(2x)^2}{2x^2+1} \times 2.
  5. Simplify Expression: Simplify the expression: F′(x)=4x24x2+1×2F'(x) = \frac{4x^2}{4x^2+1} \times 2.
  6. Final Result: F′(x)=8x24x2+1F'(x) = \frac{8x^2}{4x^2+1}.

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