Question

Use any method to find the integrals.

2. S x2 + 4dx

Answer 1

Problem 2:

Formulae to remember

1.    $\frac{d}{dx}(tanx)=sec^{2}x$

2     .$\int sec(x) dx= ln(tan(x)+sec(x))$

3.   $sec^{2}(x)-tan^{2}(x)=1$

2. Wx²+4 dx = I Let's consider x=2 tan (u) u dx = 2Sea (u). du - I l utan (w) +4 )x a Secľudu => I = | asec*(w). 2. Vittan*us du Since, 1+ tan (u) = sec (4) I= 45 sec (u) du We use the reduction formula [sec (xsdx = 0 - 2 1 sec (x) dx + Seen-2 x. tanx n-1

Sec (u) du - Sec (us. tan(u) + £ 1 sec (u du - Sec(u) tan(us + - ln ( Sectus •[ [sec (x) dx = ln( tan(25+Sec (2))] 2 + tan (u)) :. I= 4/sec (u) du - 2 Sec (u). tan(u) + ain(Sec (4) + tan (u)) Now re- substituting back uz tan-" (x2) tanu= Seck = √ + (%)²

i. Integral changes to where c of in the constant integration,

Problem 3:

Formulae to remember:

1.    $\frac{d}{dx}(arctanx)=\frac{1}{1+x^{2}}$

2. $\int \frac{1}{x}= ln|x|$

3. Integration by parts

arc (tants de = I We do integration by parts where Judo= uv-sudu where u= arc(tanx) dv = - = - 1 / 2 I@arctamed :. I = arc tanx - 1 (arctanx) - dz

Now solve for Stilitz, die 14x²-x² da all4x²) Es under best - Something - de - - 1 z 2 de 2x In(x) = 1 1 2x de Let u=ltro du=zade - Incu) = ln (142²) - In(a) = 12 ( 11 (1+2²)

Now substituting everything in I= - arctank & Inlal tc where - 1 1 1 1+x² in the constant of integration