∫xcos⁡x+12x3esin⁡x+x2dx

I=∫(xcos⁡x+1)esin⁡xdxesin⁡xx2xesin⁡x+1 Put 2xesin⁡x+1=t2⇒ 2esin⁡x+esin⁡xcos⁡xxdx=2tdt⇒ esin⁡x(1+xcos⁡x)dx=tdt∴ I=∫tdtt2−12⋅t I=2∫dtt2−1=log⁡t−1t+1+C,where t2=2xesin⁡x+1

∫xcos⁡x+12x3esin⁡x+x2dx

I=∫(xcos⁡x+1)esin⁡xdxesin⁡xx2xesin⁡x+1 Put 2xesin⁡x+1=t2⇒ 2esin⁡x+esin⁡xcos⁡xxdx=2tdt⇒ esin⁡x(1+xcos⁡x)dx=tdt∴ I=∫tdtt2−12⋅t I=2∫dtt2−1=log⁡t−1t+1+C,where t2=2xesin⁡x+1

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$\int \frac{xcos x + 1}{\sqrt{2 x^{3} e^{\sin x} + x^{2}}} dx$

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$\log |\frac{\sqrt{2 {xe}^{\sin x} + 1} - 1}{\sqrt{2 {xe}^{\sin x} + 1} + 1}| + C$

$\log |\frac{\sqrt{2 {xe}^{\sin x} - 1} + 1}{\sqrt{2 {xe}^{\sin x} + 1} + 1}| + C$

$\log |\frac{\sqrt{2 {xe}^{\sin x} + 1} + 1}{\sqrt{2 {xe}^{\sin x} - 1} + 1}| + C$

$\log |\frac{\sqrt{2 {xe}^{\sin x} + 1} + 1}{\sqrt{2 {xe}^{\sin x} - 1} - 1}| + C$

answer is A.

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Detailed Solution

$I = \int \frac{(xcos x + 1) e^{\sin x} dx}{e^{\sin x} x \sqrt{2 {xe}^{\sin x} + 1}}$

$Put 2 {xe}^{\sin x} + 1 = t^{2}$

$\begin{aligned} \Rightarrow 2 [e^{\sin x} + e^{\sin x} \cos xx] dx = 2 tdt \\ \Rightarrow e^{\sin x} (1 + xcos x) dx = tdt \\ ∴ I = \int \frac{tdt}{\frac{(t^{2} - 1)}{2} \cdot t} \\ I = 2 \int \frac{dt}{t^{2} - 1} = \log |\frac{t - 1}{t + 1}| + C, \end{aligned}$