If int frac{x^4+1}{x^2+1} dx = Ax^3 + Bx^2 + | vedclass

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Question

(B) To solve the integral $\int \frac{x^4+1}{x^2+1} dx$,we perform polynomial division or algebraic manipulation.We can write $x^4+1$ as $(x^4-1) + 2

Vedclass · Accepted Answer

$\frac{4}{3}$

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(B) To solve the integral $\int \frac{x^4+1}{x^2+1} dx$,we perform polynomial division or algebraic manipulation.We can write $x^4+1$ as $(x^4-1) + 2 = (x^2-1)(x^2+1) + 2$.Thus,$\frac{x^4+1}{x^2+1} = \frac{(x^2-1)(x^2+1) + 2}{x^2+1} = x^2 - 1 + \frac{2}{x^2+1}$.Now,integrate each term with respect to $x$:$\int (x^2 - 1 + \frac{2}{x^2+1}) dx = \int x^2 dx - \int 1 dx + 2 \int \frac{1}{x^2+1} dx$.$= \frac{x^3}{3} - x + 2 	an^{-1} x + E$.Comparing this with $Ax^3 + Bx^2 + Cx + D 	an^{-1} x + E$,we get:$A = \frac{1}{3}$,$B = 0$,$C = -1$,$D = 2$.Therefore,$A+B+C+D = \frac{1}{3} + 0 - 1 + 2 = \frac{1}{3} + 1 = \frac{4}{3}$.

If $\int \frac{x^4+1}{x^2+1} dx = Ax^3 + Bx^2 + Cx + D \tan^{-1} x + E$,then $A+B+C+D=$

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