Choose the correct answer-1- x 2-2 x -2 dx equals- a x tan -1x-1-Cbtan -1x-1-Cc x-1 tan -1 x-Cdtan -1 x - C

Let I=∫1/x^2+2x+2dx =∫1/(x+1)^2+1^2dx Let x+1=t ⇒ dx=dt ∴ I=∫1/t^2+1^2dx =1/1tan^ 1(t/1)+C [ ∵∫dx/a^2+x^2=1/atan^ 1(x/a) ] =tan^ 1 ( x+1/1 )+C =tan^ 1(x+1) ...

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Byju's Answer

Standard XII

Mathematics

Integration by Partial Fractions

Choose the co...

Question

Choose the correct answer.
$\int \frac{1}{x^{2} + 2 x + 2} d x$ equals

$(a) x t a n^{- 1} (x + 1) + C (b) t a n^{- 1} (x + 1) + C (c) (x + 1) t a n^{- 1} x + C (d) t a n^{- 1} x + C$

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Solution

Let $I = \int \frac{1}{x^{2} + 2 x + 2} d x = \int \frac{1}{(x + 1)^{2} + 1^{2}} d x Let x + 1 = t \Rightarrow d x = d t$
$∴ I = \int \frac{1}{t^{2} + 1^{2}} d x = \frac{1}{1} t a n^{- 1} (\frac{t}{1}) + C [∵ \int \frac{d x}{a^{2} + x^{2}} = \frac{1}{a} t a n^{- 1} (\frac{x}{a})] = t a n^{- 1} (\frac{x + 1}{1}) + C = t a n^{- 1} (x + 1) + C$
Hence, the option (b)is correct.

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equals

A. x tan⁻¹ (x + 1) + C

B. tan^−
1 (x + 1) + C

C. (x + 1) tan⁻¹ x + C

D. tan⁻¹ x + C

Q. The integral of

{tan}^{- 1} (x)

is -

\int \tan^{- 1} \sqrt{x} d x i s e q u a l t o (a) (x + 1) \tan^{- 1} \sqrt{x} - \sqrt{x} + C (b) x \tan^{- 1} \sqrt{x} - \sqrt{x} + C (c) \sqrt{x} - x \tan^{- 1} \sqrt{x} + C (d) \sqrt{x} - (x + 1) \tan^{- 1} \sqrt{x} + C

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\int \frac{(x^{2} - 1)}{(x^{4} + 3 x^{2} + 1) {tan}^{- 1} (x + \frac{1}{x})} d x

\int \frac{x T a n^{- 1} x}{(1 + x^{2})^{3 /_{2}}} d x =

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Choose the correct answer. ∫1x2+2x+2dx equals (a)xtan−1(x+1)+C(b)tan−1(x+1)+C(c)(x+1)tan−1x+C(d)tan−1x+C

Let I=∫1x2+2x+2dx=∫1(x+1)2+12dxLet x+1=t⇒dx=dt ∴I=∫1t2+12dx=11tan−1(t1)+C[∵∫dxa2+x2=1atan−1(xa)]=tan−1(x+11)+C=tan−1(x+1)+C Hence, the option (b)is correct.

Choose the correct answer.
$\int \frac{1}{x^{2} + 2 x + 2} d x$ equals

$(a) x t a n^{- 1} (x + 1) + C (b) t a n^{- 1} (x + 1) + C (c) (x + 1) t a n^{- 1} x + C (d) t a n^{- 1} x + C$