Question

integration of 1/1+x+x^2+x^3​

Answer 1

Step-by-step explanation:

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Answer 2

hi frnd

Note that x3+x2+x+1=(x+1)(x2+1)x3+x2+x+1=(x+1)(x2+1). We use partial fracions. So we try to find constants A,B,CA,B,C such that

4x(x+1)(x2+1)=Ax+1+Bx+Cx2+1.4x(x+1)(x2+1)=Ax+1+Bx+Cx2+1.

Bring the right-hand side to the common denominator (x+1)(x2+1)(x+1)(x2+1). The numerators must be identically equal. It follows that

4x=A(x2+1)+(Bx+C)(x+1).4x=A(x2+1)+(Bx+C)(x+1).

Set x=−1x=−1. We get −4=2A−4=2A, and therefore A=−2A=−2.

On the right, the coefficient of x2x2 is −2+B−2+B. On the left it is 

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