Question

integrate the function integration tan inverse x dx

Answer 1

Answer :

Let us take,

tan⁻¹x = z

or, x = tanz

Then, dx = sec²z dz

Since, sec²z - tan²z = 1

or, sec²z - x² = 1

or, sec²z = 1 + x²

or, secz = √(1 + x²)

Now, ∫ tan⁻¹x dx

= ∫ z sec²z dz

= z ∫ sec²z dz - ∫ { d/dz (z) ∫ sec²z dz } dz

= z tanz - ∫ tanz + c,

since ∫ sec²z dz = tanz and c is integral constant

= z tanz - log (secz) + c,

since ∫ tanz dz = log (secz)

= x tan⁻¹x - log {√(1 + x²)} + c

= x tan⁻¹x - (1/2) log (1 + x²) + c,

which is the required integral

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Answer 2

$I = \int{\tan^{ - 1} x } \, dx \\ \\ I = \int{\tan^{ - 1} x * 1 } \, dx \\ \\ I = \tan^{ - 1}* \int{1} \, dx \, \, - \, \int{ \frac{d(\tan^{ - 1}x)}{dx} \int{1} \, dx } \\ \\ I = xtan^{ - 1}x \, - \int{\frac{x}{1+ x^{ 2}}}dx \\ \\ I = xtan^{ - 1}x \, - I_1(let) \\ Now, \\ \\ I_1 = \int{\frac{x}{1 + x^{ 2}}dx$

$let, 1 +x^{2} = t \\ \\ xdx = \frac{dt}{2}\\ \\ So, \\ \\ I_1 = \frac{1}{2}\int{\frac{1}{t}\,dt \\ \\ =\frac{1}{2} logt$

$I_1 = \frac{1}{2}log(1 + x^{2})} \\ \\ Therefore, \\ \\ I = xtan^{-1}x \, - \frac{1}{2}log(1 + x^{2})} + c$