Integrals

Q: If \$\displaystyle \alpha = \int_0^{1} { \left (e^{ \left (9x + 3\tan^{-1}x \right) } \right) } \left( \dfrac{12+9x^2}{1+x^2} \right) dx \$ , where \$\tan^{-1} x \$ takes only principal values, then the value of \$ \left ( \log_e |1 +\alpha| - \dfrac{3\pi}{4} \right)\$ is

\$\displaystyle \alpha = \int_0^{1} { \left (e^{ \left (9x + 3 \tan^{-1}x \right) } \right) } \left( \dfrac{12+9x^2}{1+x^2} \right) dx \$Let \$z = 9x+3\tan^{-1} x\Rightarrow dz = \cfrac{12+9x^2}{1+x^2}dx\$\$\therefore \displaystyle \alpha = \int_0^{9+\frac{3\pi}{4}} e^z dz =e^{ 9+\frac{3\pi}{4}} -1\$\$\displaystyle\Rightarrow \log(\alpha+1) =9+\frac{3\pi}{4}\$ \$\therefore \Rightarrow \log(\alpha+1) -\frac{3\pi}{4}=9\$

Question

If $α = \int_{0}^{1} (e^{(9 x + 3 t a n^{- 1} x)}) (\frac{1 2 + 9 x ^{2}}{1 + x ^{2}}) d x$ , where $tan^{- 1} x$ takes only principal values, then the value of $(lo g_{e} ∣ 1 + α ∣ - \frac{3 π}{4})$ is

Integrals

Maths 7634 Views

If α=∫01​(e(9x+3tan−1x))(1+x212+9x2​)dx , where tan−1x takes only principal values, then the value of (loge​∣1+α∣−43π​) is

Maths
7634 Views

If $α = \int_{0}^{1} (e^{(9 x + 3 t a n^{- 1} x)}) (\frac{1 2 + 9 x ^{2}}{1 + x ^{2}}) d x$ , where $tan^{- 1} x$ takes only principal values, then the value of $(lo g_{e} ∣ 1 + α ∣ - \frac{3 π}{4})$ is