How do you find the integral of sqrt(x^2+9) dx√x2+9dx?

2 Answers
Sonnhard
Jun 2, 2018

-1/2*(t^2/2+18ln(t)-81/82t^3)+C−12⋅(t22+18ln(t)−8182t3)+C where t=sqrt(x^2+9)-xt=√x2+9−x

Explanation:

Setting
sqrt(x^2+9)=t+x√x2+9=t+x
then we get
x=(9-t^2)/(2*t)x=9−t22⋅t
and
dx=-(t^2+9)/(2t^2)dtdx=−t2+92t2dt
so we get-1/2int (t^2+9)^2/t^3dt−12∫(t2+9)2t3dt
this is
-1/2int (t+18/t+81/t^3)dt=−12∫(t+18t+81t3)dt=
-1/2*(t^2/2+18ln(t)-81/(2t^2))+C−12⋅(t22+18ln(t)−812t2)+C

maganbhai P.
Jun 2, 2018

We know that,

intsqrt(x^2+a^2)dx=x/2sqrt(x^2+a^2)+a^2/2ln|x+sqrt(x^2+a^2)|+c∫√x2+a2dx=x2√x2+a2+a22ln∣∣x+√x2+a2∣∣+c

Taking, a=3a=3 , we get

intsqrt(x^2+9)dx=x/2sqrt(x^2+9)+9/2ln|x+sqrt(x^2+9)|+c∫√x2+9dx=x2√x2+9+92ln∣∣x+√x2+9∣∣+c

Explanation:

II^(nd) MethodIIndMethod

Here,

I=sqrt(x^2+9) dx...to(1)

I=intsqrt(x^2+9)*1dx

"Using "color(blue)"Integration by Parts :"

color(blue)(intu*vdx=uintvdx-int(u'intvdx)dx

Let u=sqrt(x^2+9) and v=1

=>u'=(2x)/(2sqrt(x^2+9))=x/sqrt(x^2+9) and intvdx=x+c

So,

I=sqrt(x^2+9)*x-intx/sqrt(x^2+9)*xdx

I=xsqrt(x^2+9)-intx^2/sqrt(x^2+9)dx

I=xsqrt(x^2+9)-int((x^2+9)-9)/sqrt(x^2+9)dx

I=xsqrt(x^2+9)-int(x^2+9)/sqrt(x^2+9)dx+int9/sqrt(x^2+9)dx

I=xsqrt(x^2+9)-intsqrt(x^2+9)dx+9int1/sqrt(x^2+3^2)dx

I=xsqrt(x^2+9)-I+9ln|x+sqrt(x^2+3^2)|+Ctofrom (1)

I+I=xsqrt(x^2+9)+9ln|x+sqrt(x^2+9)|+C

2I=xsqrt(x^2+9)+9ln|x+sqrt(x^2+9)|+C

I=x/2sqrt(x^2+9)+9/2ln|x+sqrt(x^2+9)|+C

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