Question

144r2+49 -da 1. Find 144x2 +49 144x2 + 49 dx- dx *+C (145x2 +5033/2 4322" -+CO (144x2 + 49,3/2 147x? -+Co 3/2 - + 98x4 (13x2+8,972 147x (169x2+61) 1447 •dx- - 144x2 +49 os dx- 14 144x2 + 49 ·da- + 1 - 1 2. Find the indefinite integral a zona de *Cp the +Or arada?4+cr het oor HC I - dx 3.Complete the square and find ' *x?+24x + 156 dx = 12: +24x+156 + (x + I Tp2 +243 +156 - 12m|vx2+1 (8+12) +- -dx = {x?+248 + 156 – 121| vix?=24x + 156 + (x+12)| dx = x2+24x +156 +In-x2+24x +156 +(x+ x+156 + (x+12)+c 4x2+24x + 156 1 (12+24x + 156 I dx = In 1x?+24x +156 +C +24x + 136 <dx=blur? T2+28+156 dx = 2x2+24x+156 – 1 – 12m|yle?+24x+156|+C •dx x? 4. Find the definite integral 711-6 of WIT+o) - 15-6 к Т-в) -ь. -us-estir Hall JT-) -tos- (W17-6) - bas-com sua his In 11 6.11

Answer 1

V1M2+49 che NA g'=1 14 integrate by pants S&gl= fg - Iflg take, for 144x²+49 i differentiating t integrating f'= 1440 I g= V 144%2+49 33 144 K2 +49 48 = de 383 Jc? 114432449 -S fi 1447%+49 1) + 48 x? J 144x +49 A dre 343 W 1942+49 3 x3 = + 48 x=7tanu 144 12 1 - seerude seerude substitute, 42 tan? uts seen 7 / selude (x12 see? uche Usanetan (121 11270 49 tamu secu 구 dra 144x4449 +48 303 - 144x2449 48x12 . +48 8x12 / seau du Tanzu 303 S S Vistent = +98x13 cosu sinau du 303 7x7 V144x3+49 - t 48X12 7X7 do 12 313 substitute, Sinua v - Caridu=do [e-integrating Constant Sathya since, Uzane tan 1219 144x2449 = + 48x12 787 ( 1 ) to te; 313 48x13 는 (1412449 3u3 7X7 sinua te [sinſanetan (3) 12R √1446²49 48 [14463 49 +1 te 7 3u3 7n 14403 45 + 1

- 1194x2+49 N 48 v 144x2+49 490 te 313 te - 3 te 49 (14412 449 – 48x3x² 14492449 14-703 √144.63 149 (49+ +14442) 19703 1700 (49+144x2382 te, fee integrating constant] 147x3 therefore, 144 62749 de a - (144x2+4972 X4 a te 14723 osli _ du (22+4)* substitute, Uzatance see? udu » dr =2 il a sec²u du (4ta2u+y))52 seen » u= anetan (2) then then, 2 du = sedu [sin(anetan) a du = 4 Scosudu 2V + 2² 4 ti ES seen Sinlu) te = 4 al te ☆ 2 5 m ² tl L X2 8142 44 te te, = Aufq [e-integrating constant ?

R 3 : che 122+298 +156 n+12-12 au 5242964 156 an I 20+12) hy - 12 cu (x2+24 4156 522 +244 4156 (2x +24) che 12 ******** eft dhe √2+12) 12 23+24x+156 Now, s du (2x+29) de Vre²+241 +56 substitute, du x²+244 +156=4 v (2x +24) dr=du ū% de ul -1/2H te, +1 = 2 rute, [EF integrating Constant] =21x²+244 +156 tai s and, du VER112)²+(12) ² substitule, ter v=1412 2/3 253 do N = Su ~128² +12 do Joti =en 7 do = ha de en (142+1 + telee Altez 2 ener (1412)2. 2412 255 tit 12

2 = In 111 647192+12 + (19)| +62 – ln(208) 2+244 +256 + 6+12) tez [23=62- lm233] 2, C are integrating constant Now, putting the value of ☺ and ind in en (2,5 we get (24+24) de du N et- - 12 so Vr²+244 4156 (+12) 2 +12 1/2 ( 2 5 22 424 x +156 -12 en 11224244256 +62+12) te,–1263 2 tept v 22 +298 +156 -12 en 1724244256 +628121 te [caintegrating constant and C=6,-1283 4. 5x²_25 dr * 82 5 50225 first we find dre R2 integrate by pants s fg = fg - Jflg take, f= 5x2_25 gl=1 differentialing I fla ga x2 & integrating a 24 25 x dhe Vr2_25

22-25 + dr R 25 r Jul-25 Substitite, Uit 5 du t dia re S 254225 » dua te da = 22-25 x t Vui 22-25 + te, re ſu2_25 z re de la endur í tu) en vent -1 +/+e, + ln / Vice as tx/+e, ln5 ten/Vx22stolte, E., C= integrerating constant] [e=c, ln 5 (22_25 Vreas re Now from * S vrees dne M2 6 = Ehe se hezar ten/12225tale endir +6) - ln (5) - SIT (which is the 5 = required answer