If -x-x2-3-2 x1-6-x1-x3 d x-A x2-3-Btan-1x1-6-C then

The correct options are A A=3/2 B B=12 Let x =y^6 dx=6y^5dy I =∫y^6+y^4+2y/y^6(1+y^2)6y^5.dy =6 ∫y^3(y^2+1)+2/(y^2+1)dy =6∫{ y^3 +2/1+y^2}dy =6 {y^4/4+2 ...

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Byju's Answer

Standard XII

Mathematics

First Principle of Differentiation

If ∫x+x2/3+2 ...

Question

If $\int \frac{x + x^{\frac{2}{3}} + 2 x^{\frac{1}{6}}}{x (1 + x^{\frac{1}{3}})} d x = A x^{\frac{2}{3}} + B t a n^{- 1} (x^{\frac{1}{6}}) + C$ then

A = \frac{3}{2}

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B = 12

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A = \frac{2}{3}

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B = \frac{1}{6}

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Solution

The correct options are
A $A = \frac{3}{2}$

B $B = 12$

$L e t x = y^{6} d x = 6 y^{5} d y I = \int \frac{y^{6} + y^{4} + 2 y}{y^{6} (1 + y^{2})} 6 y^{5} . d y = 6 \int \frac{y^{3} (y^{2} + 1) + 2}{(y^{2} + 1)} d y = 6 \int {y^{3} + \frac{2}{1 + y^{2}}} d y = 6 {\frac{y^{4}}{4} + 2 t a n^{- 1} y} + C = \frac{3}{2} x^{\frac{2}{3}} + 12 t a n^{- 1} (x^{\frac{1}{6}}) + C$

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If ∫x+x23+2x16x(1+x13)dx=Ax23+Btan−1(x16)+C then

The correct options are A A=32 B B=12 Letx=y6dx=6y5dyI=∫y6+y4+2yy6(1+y2)6y5.dy=6∫y3(y2+1)+2(y2+1)dy=6∫{y3+21+y2}dy=6{y44+2tan−1y}+C=32x23+12tan−1(x16)+C

If $\int \frac{x + x^{\frac{2}{3}} + 2 x^{\frac{1}{6}}}{x (1 + x^{\frac{1}{3}})} d x = A x^{\frac{2}{3}} + B t a n^{- 1} (x^{\frac{1}{6}}) + C$ then