Question

evaluate (integral of)✓x⁴.dx​

Answer 1

Answer:

I=∫(x2+1)x4+1x2−1dx=∫(x+x1)x2+x211−x21dx

Substituting t=x+x1⇒dt=(1−x21)dx, we get

I=∫tt2−2dt=∫t2t2−2tdt

Again substituting y2=t2−2, we get

I=∫(y2+2)yydy=

Explanation:

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