Lf -2x2-a2-x2x2-a2dx-k-x-1-atan-1x-a-c- then k-

The correct option is C 1 ∫2x^2+a^2x^2(x^2+a^2)dx= ∫x^2+(x^2+a^2)x^2(x^2+a^2)dx = ∫1(x^2+a^2)dx+ ∫1x^2dx = ^1/_a tan^ 1(^x/_a)+( 1x)+c ...

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Geometric Interpretation of Def.Int as Limit of Sum

lf ∫2x2+a2/...

Question

lf $\int \frac{2 x^{2} + a^{2}}{x^{2} (x^{2} + a^{2})} d x = \frac{k}{x} + \frac{1}{a} t a n^{- 1} \frac{x}{a} + c$ , then k=

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Similar questions

\int \frac{1}{x^{2}} log (x^{2} + a^{2}) d x = \frac{- 1}{x} log (x^{2} + a^{2}) + k . \frac{1}{a} {tan}^{- 1} \frac{x}{a} .

k

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

A {tan}^{- 1} + B {tan}^{- 1} (\frac{x}{2}) + C

B =

\int \frac{\sqrt{x^{2} - a^{2}}}{x} d x

\int \frac{2 x^{2} + 3}{(x^{2} - 1) (x^{2} + 4)} d x = a log (\frac{x - 1}{x + 1}) + b {tan}^{- 1} (\frac{x}{2}) + c

b

\int \frac{tan x}{1 + tan x + {tan}^{2} x} d x = x - \frac{K}{\sqrt{A}} {tan}^{- 1} (\frac{K tan x + 1}{\sqrt{A}}) + C

(K, A)

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