x2+2x+10=(x+1)2+9Now, we can use the substitution:v=x+1⟹dx=dvThen, we have:I2=∫1√v2+9dvThis integral can be solved using the formula:∫1√v2+a2dv=ln|v+√v2+a2|+Cwhere a=3. Thus, we get:I2=ln|v+√v2+9|+C2=ln|(x+1)+√(x+1)2+9|+C2 Step 4: Combine ...
Integrate the following. 1.int_0^infty xe^-4x \:dx 2.int_0^1ln\:x\:dx 3.int_0^infty e^x x^2x +1 dx 4.int_0^1dx sqrt1-x^2 5.int_0^infty dx 4+x^2 Use a substitution to integrate: integral x^3/sqrt(4x^4+1) dx. ...
Evaluate the following integral: integral from 0 to 4 integral from sqrt(x) to 2 of 1/(y^3 + 1) dydx. Find the following integral: \int(x^{-3} + \sqrt{x} + 2x^{-1} + 3x^{-1/2} + 2)dx. Evaluate the indefinite integral. x^2 squareroot of (9 - x^2) dx Use integra...
\[{{S}^{\text{c}}}=\frac{m\omega }{2}\left\{ \left( {{x}^{2}}+x_{0}^{2} \right)\cot \left[ \omega \left( t-{{t}_{0}} \right) \right]-2x{{x}_{0}}\csc \left[ \omega \left( t-{{t}_{0}} \right) \right] \right\}\]. ...
Evaluate the integral of (1 - 2x)^3 dx. Evaluate the integral from 0 to 1 of xe^(-x ) dx. Evaluate the integral of x^2 ln(5x) dx. Evaluate the integral \int_0^1 2/(2x^2+3x+1) dx. Evaluate the integral from 1 to e of (1/x)*sqrt(ln(x) + 3) dx. Evaluate the ...
\int (x+1)^2dx=\int x^2+2x+1dx=\frac{1}{3}x^3+x^2+x+C 法二:换元法令t=x+1 , \frac{dt}{dx}=1\Rightarrow dt=dx,把所有的 x 都用t 来代替,于是 \int (x+1)^2dx=\int t^2 dt=\frac{1}{3}t^3=\frac{1}{3}(x+1)^3+C。 我们可以发现通过代换这样积分就简单了很多...
\int x^2} - 2\sin x + \ln x \ dx} Find indefinite integral \int (1 - 3x)^{12}dx. Find the Indefinite Integral. \int 1 \over {6 - 8xdx} Find the indefinite integral. \int \frac{dx}{(2x^{2} + 4x + 7)} Find the indefinite integral. \int \frac{dx...
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Integrate by parts using the formula( ∫ udv=uv-∫ vdu), where ( u=(ln)(x)) and ( dv=x).( d((ln)(x)(1/2x^2)-∫ 1/2x^21/xdx))Simplify.( d(((ln)(x)x^2)/2-∫ x/2dx))Since ( 1/2) is constant with respect to ( x), move ( 1/2) out of the integral.( d(...
∫ cos x dx = sin x + C ∫ sec2x dx = tan x + C ∫ csc2x dx = -cot x + C ∫ sec x (tan x) dx = sec x + C ∫ csc x ( cot x) dx = – csc x + C ∫ (1/x) dx = ln |x| + C ∫ exdx = ex+ C ...