∫ dx/√(〖2x〗^2+3x+4)

Question

pls help me for my question answer.

Answer 1

∫dx/√((2x)²+3x+4)=∫dx/(2√(x²+3x/4+1))=

∫dx/2√((x+3/8)²+1-9/64)=∫dx/2√((x+3/8)²+55/64).

Let x+3/8=(√55/8)tanθ, dx=(√55/8)sec²θdθ.

∫dx/2√((x+3/8)²+55/64)=½∫(√55/8)sec²θdθ/((√55/8)secθ)=

½∫secθdθ=½∫secθ(secθ+tanθ)dθ/(secθ+tanθ)=

½∫(secθtanθ+sec²θ)dθ/(secθ+tanθ)=

½ln(secθ+tanθ)+C where C is the constant of integration. But tanθ=(8/√55)(x+3/8)=(8x+3)/√55=(8x+3)√55/55 and secθ=√(1+(8x+3)²/55). The integral evaluates to ½ln(a(√(1+(8x+3)²/55)+(8x+3)√55/55)), where a is a constant.