The correct option is A 12+x^ 5+x^ 6+C Taking x^7 common from denomiater ∫5x^8+6x^7/x^14(2+x^ 5+x^ 6)^2 dx =∫5x^8+6x^ 7/x^14(2+x^ 5+x^ 6)^2 dx ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Dividing a Polynomial by a Monomial

∫5x8+6x72x7+x...

Question

$\int \frac{5 x^{8} + 6 x^{7}}{(2 x^{7} + x^{2} + x)^{2}} d x =$

\frac{1}{2 + x^{- 5} + x^{- 6}} + C

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{2 + x^{- 10} + x^{- 7}} + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{2 + x^{- 2} + x^{- 5}} + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{2 + x^{- 4} + x^{- 10}} + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{1}{2 + x^{- 5} + x^{- 6}} + C$
Taking $x^{7}$ common from denomiater
$\int \frac{5 x^{8} + 6 x^{7}}{x^{14} (2 + x^{- 5} + x^{- 6})^{2}} d x = \int \frac{5 x^{8} + 6 x^{- 7}}{x^{14} (2 + x^{- 5} + x^{- 6})^{2}} d x$
det $2 + x^{- 5} + x^{- 6} = t$ $- 5 x^{- 6} - c x^{- 7} d x = d t$
$\Rightarrow I = \int \frac{- d t}{t^{2}} = - \int \frac{d t}{t^{2}} = - (\frac{- 1}{t}) = \frac{1}{t} + c$
$\Rightarrow I = \frac{1}{2 + x^{- 5} + x^{- 6}} + C \Rightarrow (A)$

Suggest Corrections

Similar questions

\int \frac{x^{9} d x}{(4 x^{2} + 1)^{6}}

^{'} C^{'}

x_{1} x_{2} x_{3} x_{4} x_{5} x_{6}

x_{1} < x_{2} < x_{3} < x_{4} < x_{5} < x_{6}

The value of the integral $\int (x^{2} + x) (x^{- 8} + 2 x^{- 9})^{1 / 10} d x$ is

x_{1} x_{2} x_{3} x_{4} x_{5} x_{6}

x_{1} < x_{2} < x_{3} \leq x_{4} < x_{5} < x_{6}

\int \frac{x^{9}}{{(4 x^{2} + 1)}^{6}} d x

(a) \frac{1}{5 x} {(4 + \frac{1}{x^{2}})}^{- 5} + C (b) \frac{1}{5} {(4 + \frac{1}{x^{2}})}^{- 5} + C (c) \frac{1}{10 x} {(\frac{1}{x^{2}} + 4)}^{- 5} + C (d) \frac{1}{10} {(\frac{1}{x^{2}} + 4)}^{- 5} + C

Join BYJU'S Learning Program