Dear student, Let I=∫xdx9-16x4Put u=x22on differentiating, du=2x2dx⇒xdx=duI=-∫du64u4-9I=-∫du8u+38u-3By partial fraction-∫18u+38u-3du=-∫A8u+3du-∫B8u-3 du --[1]Solving 18u+38u-3= A8u+3+B8u-3 --[2]⇒18u-3=A+B8u+38u-3 --[multiply by 8u+3]put u=-38⇒1-3-3=A+B0-3-3⇒-16=AMultiply [2] by (8u-3)18u+3=A(8u-3)8u+3+Bput u=38⇒13+3=A8×38-38×38+3+B⇒16=Bfrom [1]-∫18u+38u-3du=16∫18u+3du-16∫18u-3 du=log8u+36×8-log8u-36×8+C=log8u+348-log8u-348+Cundo substitution u=x22I=log8x22+348-log8x22-348+CI=log4x2+348-log4x2-348+C Regards

Pls fast dj

Pls fast dj J 9 — 16x4

Dear student,

L e t I = \int \frac{x d x}{9 - 16 x^{4}} P u t u = \frac{x^{2}}{2} o n d i f f e r e n t i a t i n g, d u = \frac{2 x}{2} d x \Rightarrow x d x = d u I = - \int \frac{d u}{64 u^{4} - 9} I = - \int \frac{d u}{(8 u + 3) (8 u - 3)} B y p a r t i a l f r a c t i o n - \int \frac{1}{(8 u + 3) (8 u - 3)} d u = - \int \frac{A}{8 u + 3} d u - \int \frac{B}{8 u - 3} d u - - [1] S o l v i n g \frac{1}{(8 u + 3) (8 u - 3)} = \frac{A}{8 u + 3} + \frac{B}{8 u - 3} - - [2] \Rightarrow \frac{1}{(8 u - 3)} = A + \frac{B (8 u + 3)}{8 u - 3} - - [m u l t i p l y b y (8 u + 3)] p u t u = - \frac{3}{8} \Rightarrow \frac{1}{(- 3 - 3)} = A + \frac{B (0)}{- 3 - 3} \Rightarrow - \frac{1}{6} = A M u l t i p l y [2] b y (8 u - 3) \frac{1}{(8 u + 3)} = \frac{A (8 u - 3)}{8 u + 3} + B p u t u = \frac{3}{8} \Rightarrow \frac{1}{(3 + 3)} = \frac{A (8 \times \frac{3}{8} - 3)}{8 \times \frac{3}{8} + 3} + B \Rightarrow \frac{1}{6} = B f r o m [1] - \int \frac{1}{(8 u + 3) (8 u - 3)} d u = \frac{1}{6} \int \frac{1}{8 u + 3} d u - \frac{1}{6} \int \frac{1}{8 u - 3} d u = \frac{\log (8 u + 3)}{6 \times 8} - \frac{\log (8 u - 3)}{6 \times 8} + C = \frac{\log (8 u + 3)}{48} - \frac{\log (8 u - 3)}{48} + C u n d o s u b s t i t u t i o n u = \frac{x^{2}}{2} I = \frac{\log (\frac{8 x^{2}}{2} + 3)}{48} - \frac{\log (\frac{8 x^{2}}{2} - 3)}{48} + C I = \frac{\log (4 x^{2} + 3)}{48} - \frac{\log (4 x^{2} - 3)}{48} + C

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Pls fast dj Pls fast dj J 9 — 16x4

Pls fast dj

Pls fast dj J 9 — 16x4