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This is absolutely wrong...!haque said:this question has a lot of working and i can't put up my working because of the lack of symbols( i don't have that much time-even if i scanned it u guys won't be able to read my writing. this querstion also requires a lot of assumptions as no restrictions or conditions on the constant are specified-however i just dealt with the idea that the constants were greater than zero and thus took common factors out of square root signs without writing them as absolute values(which strictly speaking is the case: sq root x squared=abs value x.) my working probably had some careless error in it however i first took out a factor of C and completed the square in the square root bit, i assumed that m=E/C -D^2/4C^2 was greater than zero. i let u=x + D/2C and let K=B-AD/2C. after i worked through this i let u be sq root m tan@ from which it simplified down to 1/Asq rootC integral d@/sq root msin@ plus bcos @ where b=K/A. use the t=tan@/2 formula and resolve into partial fractions. after the integration is done, using the tan@ relationship from before form a quadratic in tan@/2. solving for this and ignoring the minus sign(assuming tan@/2 is positive as @/2 is acute) find tan@/2. after this u get an expression in "u". substitute u=x +D/2C or whatever and u get something like 1/Ab√mc/b<SUP>2</SUP>-c ln√m + √m+(x+D/2c)<SUP>2</SUP> –(x+D/2c)(√m/b +√m/b<SUP>2 </SUP>–1)/ √m + √m+(x+D/2c)<SUP>2</SUP>–(x+D/2c)(√m/b -√m/b<SUP>2 </SUP>–1) + K<SUB>1fficeffice" /><O
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If you can split it into partial fractions as u/(Ax+B) + w/(sqrt(Cx^2+Dx+E), then you could easily solve the integral using two standard integrals. Although those constants everywhere would be EXTREMELY messy.Riviet said:The other day, I was wondering how something like this could be integrated:
1/{(Ax+B)rt(Cx2+Dx+E)}