Question

Evaluate

1 da

Answer 1

Problem: da 1 Factor the denominator: 2 1 -es dz (x1) (a4-3 + x2 -+1) Perform partial fraction decomposition: 2a342 - 3 -/5 2 da 5 (r4- 2 - 1) 5 (1) Apply linearity: 23-42I - 3 da 3 2- 1 dz +1 Now solving: da du (stens): Substitute u = +1- da -/ du This is a standard integral: n(u) Undo substitution u = +1 In(z1) Now solving: 2r3-42a - 3 dz 1 Factor the denominator: 2r3 4x2 -3 dz (Vi+1) 5-1) +1 2 + +1 2 Perform partial fraction decomposition: 2 5-355 2/5a-35-5 dr 5(22 + (5-1 n +2) 5(22+ (-V5-1) +2) Apply linearity: 25-355 2535-5 dz + dr V5J 22+ 2m2 + (-v5-1) z 5-1)+2 5J (va 2 Now solving: 2 5-35+5 +(v5-1)+2 da 2ar2 5- /5 3/5+5 and split V5 V5-1) Write 2 /5a-3/5+5 as 2 (1-9A+ #m)sA 2x2+5-1 +2 (v-1)v5 355 da 2 2 (2e2 + (V3-1) +2) Apply linearity:

Apply linearity: (V1)/ 42r+5-1 1 - 3V5+5 dz + da (-1)+2 (5-1)+2 2 2T2+ 2r2+ Now solving: 42T+ 5-1 dz (v3-1)+2 (-1)+2 2r2+ Substitute u22+ da= du (steps): 4+5-1 - Use previous result: In(u) +(v-1)2+2 I((-1)+) 2 Undo substitution u Now solving da -1)+ 2 2r2+5 Complete the square: 1 dr (V3-1) +2 V5-1 2 4ac+5-1 da- du (steps): Substitute u = V2 (V5-1 2 12 -1) -/ -(-1) V5 16 du u2+2510 16- Simplify: 1 1 du u2 1 V2V2 Now solving du This is a standard integral: arctan(u) Plug in solved integrals:

Plug in solved integrals: 1 du u21 (V5-I 8 arctan(u) (V-1) V2V2 4+5-1 Undo substitution u- (v5-1) 4+5-1 arctan (vB-1) (V51) solved integrals: Plug (V8-1) V 4ar-+ 5-1 1 - 3/5+5 da (VE-1)+2 2 + ap 2x2 5-1 +2 22 (v-1)v 4r+v5-1 355arctan 3in(2.2 + (v5-1) z +2) (5-) 2 (V5-1) 2V2 Now solving: 2/5-35-5 dar 2 2r2 -V5 Write 2 /5-3/5-5 as 4 V5-1)+ 35-5 and split: 2 (4- 5-1 - 35 5 da )+2) V-1)a 2(2p2+(-5 222 +2 Apply linearity: -)/= V5-1 5 4x 5 1 1 - 3V5-5 dr+ 22 + (-v5=1)#+2 2 V5- 2r2 Now solving: da 2p2+(-5-1) +2 (-v Substitute u 2 5-1)+2-da= du (stens) 4-5-1 du u Use previcus result: -In(u)

Undo substitution u 2 z+(1-SA) -1)+2) -V5- Now solving: / 1 da (-v5-)2+2 2r2+ Complete the square: 1 dz 2 (-3-1) +2 (vt V2 2i2-5.1) 16-(5+1) (10-(5+1) 4-5-1 du (steps): Substitute u (Vi du u2 25+10 Simplify 1 1 du u21 (va41) 8 Now solving: 1 du u2 +1 Use previous result: arctan(u) Plug in solved integrals: 1 1 du u21 (t1) arctan(u) (v+1) V5-1 4T Undo substitution u- (v5+1) 22 8 4r--1 arctan (aly/2 (vi1) (5+1) Plug in solved integrals:

5-1)5 4x- 5-1 1 3V5-5 dr 5-1+2 2 2ar2 5 2a2 t (-5-1v3v5-5 arctan 4- 2 () 5- + 2 (v5+1) v22 Plug in solved integrals: 25355 25x 3 5-5 dr t 22+(5-1 2 da 222 5-1)r+2 (le (-51)v5 5-1 - 3/5- 5 arctan Vla(20 + (-v5 =1) = +2) () 2 (v+1) Plug in solved integrals: 2 5-355 2/5a 35- 5 1 dz t daz 2a2 +(-v5-1)a+2 5 22+ 2 - (1-2^)+ (_i In(209 + (-1) 4 2) (- 1)2) etan vavav2 Mn) Plug In 2r-42+z-3 zp (1 35 n(2 + (v-1)+2) In(2- ( -5- 1) z + 2 21n(z1) VisiaCa y visiy2- The prcbiems sobed Acphy the at s tie ens ol loganithm functiong grder to ortend 2 z+1In(2+(vš-1)+2) + (-vš -1) 2+2) vialy'a (urex) Rewrite/simplify: 1)+2) + In (2 (2 1)+2) -4 In( |x + 1|) + 1 5-1 - 5 - +C 10