Let Sx-d x-ex-8 e-x-4 e-3 x- Rx-d x-ex-8 ex-4 e-x and Mx-Sx-2 Rx- If Mx-1-2tan -1fx-c then f0-

The correct option is A 3/2M(x)=∫e^x (e^2x 2)/e^4x+8e^2x+4dx put e^x=t⇒∫(t^2 2)dt/t^4+8t^2+4 =1/2tan^ 1 ( t+2/t/2 )+c =1/2tan^ 1 ( e^x+2e^ x/2 )+c ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

Let Sx=∫d x/e...

Question

Let $S (x) = \int \frac{d x}{e^{x} + 8 e^{- x} + 4 e^{- 3 x}}, R (x) = \int \frac{d x}{e^{3 x} + 8 e^{x} + 4 e^{- x}}$ and $M (x) = S (x) - 2 R (x) .$ If $M (x) = \frac{1}{2} t a n^{- 1} (f (x)) + c$ then f(0)=

\frac{3}{2}

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\frac{1}{2}

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\frac{5}{2}

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\frac{7}{2}

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Solution

The correct option is A $\frac{3}{2}$
$M (x) = \int \frac{e^{x} (e^{2 x} - 2)}{e^{4 x} + 8 e^{2 x} + 4} d x p u t e^{x} = t \Rightarrow \int \frac{(t^{2} - 2) d t}{t^{4} + 8 t^{2} + 4} = \frac{1}{2} t a n^{- 1} (\frac{t + \frac{2}{t}}{2}) + c = \frac{1}{2} t a n^{- 1} (\frac{e^{x} + 2 e^{- x}}{2}) + c$

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Similar questions

Q. Let

S (x) = \int \frac{d x}{e^{x} + 8 e^{- x} + 4 e^{- 3 x}}, R (x) = \int \frac{d x}{e^{3 x} + 8 e^{x} + 4 e^{- x}}

and

M (x) = S (x) - 2 R (x) .

M (x) = \frac{1}{2} t a n^{- 1} (f (x)) + c

then f(0)=

Q. Let

I_{1} = \int \frac{d x}{e^{x} + 8 e^{- x} + 4 e^{- 3 x}}

and

I_{2} = \int \frac{d x}{e^{3 x} + 8 e^{x} + 4 e^{- x}}

, then value of

I = I_{1} - 2 I_{2}

is equal to

Q. Let

f (x) = \frac{1}{e^{x} + 8 e^{- x} + 4 e^{- 3 x}}

and

g (x) = \frac{1}{e^{3 x} + 8 e^{x} + 4 e^{- x}} .

\int (f (x) - 2 g (x)) d x = h (x) + C,

where

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lim x \to \infty h (x) = \frac{π}{4},

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Q. If

f (x) = e^{x} + \int_{0}^{1} (e^{x} + t e^{- x}) f (t) d t

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\int \frac{3 e^{x} - 5 e^{- x}}{4 e^{x} + 5 e^{- x}} d x = a x + b ln (4 e^{x} + 5 e^{- x}) + C

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Let S(x)=∫dxex+8e−x+4e−3x, R(x)=∫dxe3x+8ex+4e−x and M(x)=S(x)−2R(x). If M(x)=12tan−1(f(x))+c then f(0)=

The correct option is A 32M(x)=∫ex(e2x−2)e4x+8e2x+4dxput ex=t⇒∫(t2−2)dtt4+8t2+4=12tan−1(t+2t2)+c=12tan−1(ex+2e−x2)+c

Let $S (x) = \int \frac{d x}{e^{x} + 8 e^{- x} + 4 e^{- 3 x}}, R (x) = \int \frac{d x}{e^{3 x} + 8 e^{x} + 4 e^{- x}}$ and $M (x) = S (x) - 2 R (x) .$ If $M (x) = \frac{1}{2} t a n^{- 1} (f (x)) + c$ then f(0)=