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Question Number 73155 by MJS last updated on 06/Nov/19
repostingaformerquestion...∫x5−1x+1dx=[t=x10→dx=10x910dx]=10∫t9(t−1)t4−t3+t2−t+1dt==10∫(t6−t4−t)dt+10∫t(t2−t+1)t4−t3+t2−t+1dt==107t7−2t5−5t2+(5+5)∫tt2−1−53t+1dt+(5−5)∫tt2−1+52t+1dt=andit′seasytosolvethese
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![reposting a former question... ∫(((x)^(1/5) −1)/((√x)+1))dx= [t=(x)^(1/(10)) → dx=10(x^9 )^(1/(10)) dx] =10∫((t^9 (t−1))/(t^4 −t^3 +t^2 −t+1))dt= =10∫(t^6 −t^4 −t)dt+10∫((t(t^2 −t+1))/(t^4 −t^3 +t^2 −t+1))dt= =((10)/7)t^7 −2t^5 −5t^2 +(5+(√5))∫(t/(t^2 −((1−(√5))/3)t+1))dt+(5−(√5))∫(t/(t^2 −((1+(√5))/2)t+1))dt= and it′s easy to solve these](Q73155.png)