∫ (x8-9 x2+18/x4-3 x2+3) d x=

Question

∫ (x8-9 x2+18/x4-3 x2+3) d x= (A) (x4/4)+x3+6 x2+c (B) (x5/5)+(x4/4)+6 x+c (C) (x5/5)+x3+6 x+c (D) (x5/5)-(x3/2)+6 x2+c

Tardigrade Admin · Accepted Answer

∫ (x8-9 x2+18/x4-3 x2+3) d x ∵ (x8-9 x2+18/x4-3 x2+3)-x4+3 x2+(6 x4-18 x2+18/x4-3 x2+3) ⇒ x4+3 x2+6 Now, ∫ x4+3 x2+6 d x ⇒ (x5/5)+(3 x3/3)+6 x+c ⇒ (x5/5)+x3+6 x+c

Q. $\int \frac{x ^{8} - 9 x ^{2} + 18}{x ^{4} - 3 x ^{2} + 3} d x =$

$\frac{x ^{4}}{4} + x^{3} + 6 x^{2} + c$

$\frac{x ^{5}}{5} + \frac{x ^{4}}{4} + 6 x + c$

$\frac{x ^{5}}{5} + x^{3} + 6 x + c$

$\frac{x ^{5}}{5} - \frac{x ^{3}}{2} + 6 x^{2} + c$

Q. ∫x4−3x2+3x8−9x2+18​dx=

4x4​+x3+6x2+c

5x5​+4x4​+6x+c

5x5​+x3+6x+c

5x5​−2x3​+6x2+c

∫x4−3x2+3x8−9x2+18​dx ∵x4−3x2+3x8−9x2+18​−x4+3x2+x4−3x2+36x4−18x2+18​ ⇒x4+3x2+6 Now, ∫x4+3x2+6dx ⇒5x5​+33x3​+6x+c ⇒5x5​+x3+6x+c

Q. $\int \frac{x ^{8} - 9 x ^{2} + 18}{x ^{4} - 3 x ^{2} + 3} d x =$

$\frac{x ^{4}}{4} + x^{3} + 6 x^{2} + c$

$\frac{x ^{5}}{5} + \frac{x ^{4}}{4} + 6 x + c$

$\frac{x ^{5}}{5} + x^{3} + 6 x + c$

$\frac{x ^{5}}{5} - \frac{x ^{3}}{2} + 6 x^{2} + c$