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# Indefinite Integrals

Please see the attachment for the questions.

Please solve each problem step by step giving solutions please. SHOW every step getting to the answer. Show substitutions, etc. DO NOT SKIP STEPS PLEASE! Look below for attachments. Adult student asking for help and I learn by the examples you solve. I learn different than others and please respect the way I learn. Thanks!

If you cannot read the exponents please make an exponent up that may look like the one in the problem. Just tell me you changed the exponent.

Please do problems 1 through 6 (both parts a and b).

Please do Problems 7 -37 odd.

Problem 16
Problem 38

Problems 39 through 47odd

#### Solution Preview

Please see the attached files for the complete solutions.

All these problems involve the principle of writing a given rational expression in partial fractions, in which the degree in numerator is less than that of denominator.

1 (a)
Write 2x=A(x+3)+B(3x+1)
Put x=0 on both sides, which gives, 3A+B=0
Put x=1 on both sides, which gives, 4A+4B=2. Solving both the equations we get, A=-1/4 and B=3/4.

So,
Plugging the values of A and B, we get the integral as

1 (b)
Write 1=Ax+Bx(x+1)+C(x+1)2
Put x=0 on both sides, which gives, C=1
Put x=-1 on both sides, which gives, A=-1
Put x=1 on both sides, which gives, A+2B+4C=1 that gives on simplification, B=-1

Hence we have 1=-x-x(x+1)+(x+1)2
So,

2(a)
Write x-1= A(x+1)+Bx(x+1)+Cx2
Put x=0 on both sides, which gives, A=-1
Put x=-1 on both sides, which gives, C=-2
Put x=1 on both sides, which gives, 2A+2B+C=0 that gives on simplification, B=2

Hence we have x-1= -(x+1)+2x(x+1)-2x2

2(b)
Write x-1= x(Ax+B)+C(x2+1)
Put x=0 on both sides, which gives, C=-1
Put x=1 on both sides, which gives, A+B+2C=0 which ...

#### Solution Summary

Around 25 integrals (definite and indefinite) are evaluated using substitution method, partial fractions method. Some of them need good concept knowledge and skill to compute.

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