Find -3 x4-4 x3-x4-x-12 d x

The correct option is A x^4/x^4+x+1+CI=∫3x^4+4x^3/(x^4+x+1)^2dx =∫3x^ 4+4x^ 5/(1+x^ 3+x^ 4)^2dx Put (1+x^ 3+x^ 4)=t ⇒ ( 3x^ 4 4x^ 5)dx=dt ⇒ I= ∫dt/t^2=1/t+C=x ...

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

Byju's Answer

Standard XII

Mathematics

Standard Formulae - 2

Find ∫3 x4+4 ...

Question

$Find \int \frac{3 x^{4} + 4 x^{3}}{(x^{4} + x + 1)^{2}} d x$

\frac{x^{4}}{x^{4} + x + 1} + C

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

\frac{1}{x^{4} + x + 1} + C

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

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

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

\frac{1}{2 (x^{4} + x + 1)^{3}} + C

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

Open in App

Solution

The correct option is A $\frac{x^{4}}{x^{4} + x + 1} + C$
$I = \int \frac{3 x^{4} + 4 x^{3}}{(x^{4} + x + 1)^{2}} d x = \int \frac{3 x^{- 4} + 4 x^{- 5}}{(1 + x^{- 3} + x^{- 4})^{2}} d x Put (1 + x^{- 3} + x^{- 4}) = t \Rightarrow (- 3 x^{- 4} - 4 x^{- 5}) d x = d t \Rightarrow I = - \int \frac{d t}{t^{2}} = \frac{1}{t} + C = \frac{x^{4}}{x^{4} + x + 1} + C$

Suggest Corrections

Similar questions

Find the quotient when $3 x^{4} - 4 x^{3} - 3 x - 1$ divided by x – 1

Q. Find

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

Q. Find the remainder when

3 x^{4} - 4 x^{3} - 3 x - 1

is dividend by

x - 1

Using Remainder theorem, find the remainder when 3x⁴ – 4x³ – 3x – 1 by x - 1.

\int_{0}^{1} \frac{(3 x^{4} + 4 x^{3} + 3 x^{2})}{(4 x^{3} + 3 x^{2} + 2 x + 1)^{2}} d x

is equal to

Join BYJU'S Learning Program

Find ∫3x4+4x3(x4+x+1)2dx

The correct option is A x4x4+x+1+CI=∫3x4+4x3(x4+x+1)2dx=∫3x−4+4x−5(1+x−3+x−4)2dxPut (1+x−3+x−4)=t⇒(−3x−4−4x−5)dx=dt⇒I=−∫dtt2=1t+C=x4x4+x+1+C

$Find \int \frac{3 x^{4} + 4 x^{3}}{(x^{4} + x + 1)^{2}} d x$