displaystyle ∫ (x2 - 2/x4 + 4)dx is equal to (where C is constant of integration)

Question

displaystyle ∫ (x2 - 2/x4 + 4)dx is equal to (where C is constant of integration) (A) (1/4)log|(x2 + 2 x + 2/x2 - 2 x + 2)|+C (B) (1/4)log|(x2 - 2 x + 2/x2 + 2 x + 2)|+C (C) (1/2)(tan)- 1((x2 + 2 x + 2/x2 - 2 x + 2))+C (D) (1/2)(tan)- 1((x2 - 2 x + 2/x2 + 2 x + 2))+C

Tardigrade Admin · Accepted Answer

displaystyle ∫ (x2 - 2/x4 + 4)dx⇒ displaystyle ∫ (1 - (2/x2)/x2 + (4)x2)dx displaystyle ∫ (1 - (2/x2)/(x + (2)x)2 - 4)dx Put (x + (2/x))=t displaystyle ∫ (d t/t2 - 4) (1/4)log|(t - 2/t + 2)|+C=(1/4)log|(x2 - 2 x + 2/x2 + 2 x + 2)|+C

Q. $\int \frac{x ^{2} - 2}{x ^{4} + 4} d x$ is equal to (where $C$ is constant of integration)

$\frac{1}{4} l o g ∣ ∣ \frac{x ^{2} + 2 x + 2}{x ^{2} - 2 x + 2} ∣ ∣ + C$

$\frac{1}{4} l o g ∣ ∣ \frac{x ^{2} - 2 x + 2}{x ^{2} + 2 x + 2} ∣ ∣ + C$

$\frac{1}{2} (t an)^{- 1} (\frac{x ^{2} + 2 x + 2}{x ^{2} - 2 x + 2}) + C$

$\frac{1}{2} (t an)^{- 1} (\frac{x ^{2} - 2 x + 2}{x ^{2} + 2 x + 2}) + C$

Q. ∫x4+4x2−2​dx is equal to (where C is constant of integration)

41​log∣∣​x2−2x+2x2+2x+2​∣∣​+C

41​log∣∣​x2+2x+2x2−2x+2​∣∣​+C

21​(tan)−1(x2−2x+2x2+2x+2​)+C

21​(tan)−1(x2+2x+2x2−2x+2​)+C

∫x4+4x2−2​dx⇒∫x2+x24​1−x22​​dx ∫(x+x2​)2−41−x22​​dx Put (x+x2​)=t ∫t2−4dt​ 41​log∣∣​t+2t−2​∣∣​+C=41​log∣∣​x2+2x+2x2−2x+2​∣∣​+C

Q. $\int \frac{x ^{2} - 2}{x ^{4} + 4} d x$ is equal to (where $C$ is constant of integration)

$\frac{1}{4} l o g ∣ ∣ \frac{x ^{2} + 2 x + 2}{x ^{2} - 2 x + 2} ∣ ∣ + C$

$\frac{1}{4} l o g ∣ ∣ \frac{x ^{2} - 2 x + 2}{x ^{2} + 2 x + 2} ∣ ∣ + C$

$\frac{1}{2} (t an)^{- 1} (\frac{x ^{2} + 2 x + 2}{x ^{2} - 2 x + 2}) + C$

$\frac{1}{2} (t an)^{- 1} (\frac{x ^{2} - 2 x + 2}{x ^{2} + 2 x + 2}) + C$