Anonymous

# integrate 2t/(1+(t^2)) dt?

2t/(1+(t^2)) dt

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• Puggy
Lv 7

∫ ( (2t)/(1 + t²) dt )

To integrate, this, use substitution. To make this clear, I'm going to rearrange the expression within the integral.

∫ ( 1/(1 + t²) 2t dt )

Let u = 1 + t². Then

du = 2t dt

Note that 2t dt is the tail end of our integral, so it follows that after the substitution, du will be our tail end.

∫ ( 1/u du )

Which is now trivial to integrate.

ln|u| + C

But u = 1 + t², so our final answer is

ln|1 + t²| + C

• Anonymous

Since the top of the fraction is the differential of the bottom then the integral is simply ln(1 + t^2) + c.

• Anonymous

put t = tan x.

then, dt = sec^2 (x) dx.

so, the integral becomes,

(2 * tan x) / (1 + tan^2 (x)) sec^2 (x) dx.

or, (2 * tan x) / (sec^2 (x)) * sec^2 (x) dx

or, 2 * tan x dx

whose integral is, - 2 * ln (cos x).

now, cos x = sq. root of cos^2 (x) = sq. root of (1 / sec^2 (x)) = sq. root of (1 / (1 + tan^2 (x))) = sq. root of (1/(1 + t^2)).

So, the answer is - 2 * ln (root of (1/(1 + t^2))) = -2 * (- 1/2) * ln (1 + t^2) = ln (1 + t^2).

• Anonymous
4 years ago

combine. ?[2sint*?(cos²t + a million)] dt allow cost = sinh? -sint dt = cosh? d? dt = (-cosh?/sint) d? ? = arcsinh[cost] cosh? = ?(sinh²? + a million) = ?(cos²t + a million) ___________ = 2?[sint*?(sinh²? + a million)] [(-cosh?/sint) d?] = -2?[cosh?*?(cosh²?)] d? = -2?cosh²? d? = -?[a million + cosh(2?)] d? = -? - (a million/2)sinh(2?) + C = -arcsinh[cost] - sinh?*cosh? + C = -arcsinh[cost] - cost*?(cos²t + a million) + C