Problem 1: Differentiation If y=(x2+1)sqrt(x)e2x, x1, find dy/dx. Problem 2: Integration Evaluate…

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Problem 1: Differentiation If y=(x2+1)sqrt(x)e2x, x>1, find dy/dx. Problem 2: Integration Evaluate I=integral(x2 x)/((1+x2)2)dx.

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Here are the solutions to your problems.

: Differentiation
Given $y = \frac{(x^2+1)\sqrt{\ln x}}{e^{3x}}$ , find $\frac{dy}{dx}$ .

Take the natural logarithm of both sides and simplify using logarithm properties.
$\ln y = \ln \left( \frac{(x^2+1)\sqrt{\ln x}}{e^{3x}} \right)$ $\ln y = \ln(x^2+1) + \ln(\sqrt{\ln x}) - \ln(e^{3x})$ $\ln y = \ln(x^2+1) + \frac{1}{2}\ln(\ln x) - 3x$
Differentiate both sides with respect to $x$ .
$\frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}(\ln(x^2+1)) + \frac{d}{dx}\left(\frac{1}{2}\ln(\ln x)\right) - \frac{d}{dx}(3x)$ $\frac{1}{y}\frac{dy}{dx} = \frac{2x}{x^2+1} + \frac{1}{2} \cdot \frac{1}{\ln x} \cdot \frac{1}{x} - 3$ $\frac{1}{y}\frac{dy}{dx} = \frac{2x}{x^2+1} + \frac{1}{2x\ln x} - 3$
Solve for $\frac{dy}{dx}$ by multiplying by $y$ and substituting the original expression for $y$ .
$\frac{dy}{dx} = y \left( \frac{2x}{x^2+1} + \frac{1}{2x\ln x} - 3 \right)$ $\frac{dy}{dx} = \frac{(x^2+1)\sqrt{\ln x}}{e^{3x}} \left( \frac{2x}{x^2+1} + \frac{1}{2x\ln x} - 3 \right)$
The final answer is $\boxed{\frac{dy}{dx} = \frac{(x^2+1)\sqrt{\ln x}}{e^{3x}} \left( \frac{2x}{x^2+1} + \frac{1}{2x\ln x} - 3 \right)}$ .

: Integration
Evaluate $I = \int \frac{x^3 \ln x}{(1+x^4)^2} dx$ .
Use integration by parts, $\int u \, dv = uv - \int v \, du$ .
Let $u = \ln x$ and $dv = \frac{x^3}{(1+x^4)^2} dx$ .
Calculate $du$ and $v$ .
$du = \frac{1}{x} dx$
To find $v$ , integrate $dv$ :
$v = \int \frac{x^3}{(1+x^4)^2} dx$
Let $w = 1+x^4$ , so $dw = 4x^3 dx \implies x^3 dx = \frac{1}{4} dw$ .
$v = \int \frac{1}{w^2} \cdot \frac{1}{4} dw = \frac{1}{4} \int w^{-2} dw = \frac{1}{4} \left( \frac{w^{-1}}{-1} \right) = -\frac{1}{4w}$
Substitute $w = 1+x^4$ back:
$v = -\frac{1}{4(1+x^4)}$
Apply the integration by parts formula.
$I = (\ln x) \left(-\frac{1}{4(1+x^4)}\right) - \int \left(-\frac{1}{4(1+x^4)}\right) \left(\frac{1}{x} dx\right)$ $I = -\frac{\ln x}{4(1+x^4)} + \frac{1}{4} \int \frac{1}{x(1+x^4)} dx$
Evaluate the remaining integral $\int \frac{1}{x(1+x^4)} dx$ .
Multiply the numerator and denominator by $x^3$ :
$\int \frac{x^3}{x^4(1+x^4)} dx$
Let $t = x^4$ , so $dt = 4x^3 dx \implies x^3 dx = \frac{1}{4} dt$ .
The integral becomes:
$\frac{1}{4} \int \frac{1}{t(1+t)} dt$
Use partial fractions for $\frac{1}{t(1+t)}$ :
$\frac{1}{t(1+t)} = \frac{A}{t} + \frac{B}{1+t}$ $1 = A(1+t) + Bt$
Setting $t=0 \implies A=1$ . Setting $t=-1 \implies B=-1$ .
So, $\frac{1}{t(1+t)} = \frac{1}{t} - \frac{1}{1+t}$ .
$\frac{1}{4} \int \left( \frac{1}{t} - \frac{1}{1+t} \right) dt = \frac{1}{4} (\ln|t| - \ln|1+t|) + C' = \frac{1}{4} \ln\left|\frac{t}{1+t}\right| + C'$
Substitute $t=x^4$ back (assuming $x>0$ ):
$\frac{1}{4} \ln\left(\frac{x^4}{1+x^4}\right) + C'$
Combine the results to get the final integral.
$I = -\frac{\ln x}{4(1+x^4)} + \frac{1}{4} \left( \frac{1}{4} \ln\left(\frac{x^4}{1+x^4}\right) \right) + C$ $I = -\frac{\ln x}{4(1+x^4)} + \frac{1}{16} \ln\left(\frac{x^4}{1+x^4}\right) + C$
The final answer is $\boxed{I = -\frac{\ln x}{4(1+x^4)} + \frac{1}{16} \ln\left(\frac{x^4}{1+x^4}\right) + C}$ .

Problem 1: Differentiation If y=(x2+1)sqrt(x)e2x, x>1, find (dy)/(dx). Problem 2: Integration Evaluate I=integral(x2 x)/((1+x2)2)dx.

Step-by-step solution

Take the natural logarithm of both sides and simplify using logarithm properties.

Differentiate both sides with respect to $x$ .

Solve for $\frac{dy}{dx}$ by multiplying by $y$ and substituting the original expression for $y$ .

Use integration by parts, $\int u \, dv = uv - \int v \, du$ .

Calculate $du$ and $v$ .

Apply the integration by parts formula.

Evaluate the remaining integral $\int \frac{1}{x(1+x^4)} dx$ .

Combine the results to get the final integral.

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