Question

10. Stan-lx dxintegration of tan inverse x ​

Answer 1

Answer:

x tan⁻¹x -1/2 ㏑(1+x²) + c

Step-by-step explanation:

Here we shall be using integration by parts

we know that, tan⁻¹x = (1)(tan⁻¹x)

Use the ILATE (Inverse,Logarithmic,Algebraic,Trigonometric,Exponential) rule to determine first and second function.

Since tan⁻¹x is an inverse function it is of higher priority and hence is first function,while 1 is an algebraic function and hence it is a lower priority second function.

Formula for integration by parts,

∫(I)(II)dx =  I∫II dx - ∫(d/dx(I))∫(II)dx)dx

Thus here, I = tan⁻¹x and II= 1

Hence integral is,

∫(tan⁻¹x)1 dx = tan⁻¹x∫1 dx - ∫(d/dx(tan⁻¹x)∫1 dx)dx

= x tan⁻¹x -∫ $\frac{1}{1+x\²}$(x) + c          

{Since we know,    d/dx(tan⁻¹x)=1/(1+x²)}

Now, we make the substitution 1+x²=t

Thus, 2x dx = dt

or x dx = (1/2) dt

Integral becomes,

x tan⁻¹x - 1/2∫dt/t + c

= x tan⁻¹x -1/2㏑t + c

= x tan⁻¹x -1/2 ㏑(1+x²) + c

Where c is the constant of integration.

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