ज्ञात करें `int(e^("m tan"^(-1)x))/((1+x^2)^("3/2"))dx`

Explanation

`z = tan^(-1)x` रखे, तो `dz=1/(1+x^2)dx` तथा `x = tan z`
अब, `I = inte^(mz)/sqrt(1+tan^2 z)dz=int e^(dz)" cos z dz"`
`e^(mz)/mcosz -int e^(xz)/m(-sinz)dz=e^(mz)/m cosz+1/"m"inte^(mz)sin z dz`
`=e^(mz)/mcos z +1/m[e^(mz)/m"sin z dz"]`
`=e^(mz)/m cos z +e^(mxz)/m^2sinz-1/m^2 I`
या, `(1+1/m^2)I=e^(mz)/m^2 (" m cos z + sin z")" ":." "I = (e^(mz)("m cos z + sin z"))/(m^(2)+1)+c`
या, `I = e^(" m tan"^(-1)x)/(m^(2)+1)(m1/sqrt(x^2+1)+x/sqrt(x^2+1))+c=(e^("m tan"^(-1)x))/((m^2+1)sqrt(x^2+1))+c`